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I 

NEWCOMB'S MATHEMATICAL COURSE 

ELEMENTS 

OF THE 

DIFFERENTIAL AND INTEGRAL 

CALCULUS 

r 

SIMON NEWCOMB 

Professor of Mathematics in the Johns Hopkins University 



OCT 31 I887J ) 

^ ■ - • . - « 



NEW YORK 
HENRY HOLT AND COMPANY 

1887 










; : 



Copyright, 1887, 

BY 

HENRY HOLT & CO. 



Ill 



PREFACE. 



The present work is intended to contain about as much 
of the Calculus as an undergraduate student, either in Arts 
or Science, can be expected to master during his regular 
course. He may find more exercises than he has time to 
work out; in this case it is suggested that he only work 
enough to show that he understands the principles they are 
designed to elucidate. 

The most difficult question which arises in treating the 
subject is how the first principles should be presented to the 
mind of the beginner. The author has deemed it best to be- 
gin by laying down the logical basis on which the whole 
superstructure must ultimately rest. It is now well under- 
stood that the method of limits forms the only rigorous basis 
for the infinitesimal calculus, and that infinitesimals can be 
used with logical rigor only when based on this method, that 
is, when considered as quantities approaching zero as their 
limit. When thus defined, no logical difficulty arises in their 
use; they flow naturally from the conception of limits, and 
they are therefore introduced at an early stage in the present 
work. 

The fundamental principles on which the use of infinitesi- 
mals is based are laid down in the second chapter. But it is 
not to be expected that a beginner will fully grasp these prin- 
ciples until he has become familiar with the mechanical pro- 
cess of differentiation, and with the application of the calcu- 



IV PREFACE. 

lus to special problems. It may therefore be found best to 
begin with a single careful reading of the chapter, and after- 
ward to use it for reference as the student finds occasion to 
apply the principles laid down in it. 

The author is indebted to several friends for advice and 
assistance in the final revision of the work. Professor John 
E. Clark of the Sheffield Scientific School and Dr. Fabian 
Franklin of the Johns Hopkins University supplied sugges- 
tions and criticisms which proved very helpful in putting the 
first three chapters into shape. Miss E. P. Brown of Wash- 
ington has read all the proofs, solving most of the problems as 
she went along in order to test their suitability. 



CONTENTS. 



PAET I. 
THE DIFFERENTIAL CALCULUS. 

PAGE 

Chapter I. Of Variables and Functions 3 

§ 1. Nature of Functions. 2. Their Classification. 3. Func- 
tional Notation. 4. Functions of Several Variables. 5. Func- 
tions of Functions. 6. Product of the First n Numbers. 7. Bi- 
nomial Coefficients. 8. Graphic Representation of Functions. 
9. Continuity and Discontinuity of Functions. 10. Many- valued 
Functions. 

Chapter II. Of Limits and Infinitesimals 17 

§ 11. Limits. 12. Infinites and Infinitesimals. 13. Properties. 
14. Orders of Infinitesimals. 15. Orders of Infinites. 

Chapter III. Of Differentials and Derivatives 25 

§ 16. Increments of Variables. 17. First Idea of Differentials 
and Derivatives. 18. Illustrations. 19. Illustration by Velocities. 
20. Geometrical Illustration. 

Chapter IV. Differentiation of Explicit Functions 31 

§ 21. The Process of Differentiation in General. 22. Differen- 
tials of Sums. 23. Differential of a Multiple. 24. Differential of 
a Constant. 25. Differentials of Products and Powers. 28. Dif- 
ferential of a Quotient of Two Variables. 27. Differentials of Ir- 
rational Expressions. 28. Logarithmic Functions. 29. Expo- 
nential Functions. 30. The Trigonometric Functions. 31. Cir- 
cular Functions. 32. Logarithmic Differentiation. 33. Velocity 
or Derivative with Respect to the Time. 

Chapter V. Functions of Several Variables and Impli- 
cit Functions 54 

§ 34. Partial Differentials and Derivatives. 35. Total Differen- 
tials. 36. Principles involved in Partial Differentiation. 37. Dif- 



vi CONTENTS. 

PAGE 

f erentiation of Implicit Functions. 38. Implicit Functions of Sev- 
eral Variables. 39. Case of Implicit Functions expressed by- 
Simultaneous Equations. 40. Functions of Functions. 41. Func- 
tions of Variables, some of which are Functions of the Others. 
42. Extension of the Principle. 43. Nomenclature of Partial 
Derivatives. 44. Dependence of the Derivative upon the Form 
of the Function. 

Chapter VI. Derivatives of Higher Orders 74 

§45. Second Derivatives. 46. Derivatives of Any Order. 
47. Special Forms of Derivatives of Circular and Exponential 
Functions. 48. Successive Derivatives of an Implicit Function. 
49. Successive Derivatives of a Product. 50. Successive Deriva- 
tives with Eespect to Several Equicrescent Variables. 51. Result 
of Successive Differentiations independent of the Order of the 
Differentiations. 52. Notation for Powers of a Differential or 
Derivative. 

Chapter VII. Special Cases op Successive Derivatives. . . 86 
§ 53. Successive Derivatives of a Power of a Derivative. 54. De- 
rivatives of Functions of Functions. 55. Change of the Equicres- 
cent Variable. 56. Two Variables connected by a Third. 

Chapter VIII. Developments in Series 95 

§ 57. Classification of Series. 58. Convergence and Divergence 
of Series. 59. Maclaurin's Theorem. 60. Ratio of the Circum- 
ference of a Circle to its Diameter. 61. Use of Symbolic Nota- 
tion for Derivatives. 62. Taylor's Theorem. 63. Identity of 
Taylor's and Maclaurin's Theorems. 64. Cases of Failure of 
Taylor's and Maclaurin's Theorems. 65. Extension of Taylor's 
Theorem to Functions of Several Variables. 66. Hyperbolic 
Functions. 

Chapter IX. Maxima and Minima of Functions of a Sin- 
gle Variable 117 

§67. Definition of Maximum Value and Minimum Value. 
68. Method of finding Maximum and Minimum Values of a Func- 
tion. 69. Case when the Function which is to be a Maximum or 
Minimum is expressed as a Function of Two or More Variables 
connected by Equations of Condition. 

Chapter X. Indeterminate Forms 128 

§70. Examples of Indeterminate Forms. 71. Evaluation of 

the Form— . 72. Forms — and X ®. 73. Form go — go. 

00 

75. Forms 0° and oo °. 



CONTENTS. Vll 

PAGE 

Chapter XI. Of Plane Curves 137 

§ 76. Forms of the Equations of Curves. 77. Infinitesimal Ele- 
ments of Curves. 78. Properties of Infinitesimal Arcs and 
Chords. 79. Expressions for Elements of Curves. 80. Equa- 
tions of Certain Noteworthy Curves. The Cycloid. 81. The 
Lenmiscate. 82. The Archimedean Spiral. 83. The Logarith- 
mic Spiral. 

Chapter XII. Tangents and Normals 147 

§84. Tangent and Normal compared with Subtangent and 
Subnormal. 85. General Equation for a Tangent. 86. Sub- 
tangent and Subnormal. 87. Modified Forms of the Equation. 
88. Tangents and Normals to the CoDic Sections. 89. Length of 
the Perpendicular from the Origin upon a Tangent or Normal. 
90. Tangent and Normal in Polar Co-ordinates. 91. Perpendicular 
from the Pole upon the Tangent or Normal. 92. Equation of 
Tangent and Normal derived from Polar Equation of the Curve. 

Chapter XIII. Of Asymptotes, Singular Points and 
Curve-tracing 157 

§93. Asymptotes. 94. Examples of Asymptotes. 95. Points 
of Inflection. 96. Singular Points of Curves. 97. Condition of 
Singular Points. 98. Examples of Double points. 99. Curve- 
tracing. 

Chapter XIV. Theory of Envelopes 169 

§100. Envelope of a Family of Lines. 101. All Lines of a 
Family tangent to the Envelope. 102. Examples and Applications. 

Chapter XY. Of Curvature, Evolutes and Involutes 180 

§103. Position; Direction; Curvature. 104. Contacts of Differ- 
ent Orders. 105. Intersection or Non-intersection of Curves ac- 
cording to the Order of Contact. 106. Radius of Curvature. 
107. The Osculating Circle. 108. Radius of Curvature when the 
Abscissa is not taken as the Independent Variable. 109. Ra- 
dius of Curvature of a Curve referred to Polar Co-ordinates. 
110. Evolutes and Involutes. 111. Case of an Auxiliary Variable. 
112. The Evolute of the Parabola. 113. Evolute of the Ellipse. 
114. Evolute of the Cycloid. 115. Fundamental Properties of 
the Evolute. 116. Involutes. 



Till CONTENTS. 

PAET II. 
THE INTEGRAL CALCULUS. 

PAGE 

Chapter I. The Elementary Forms of Integration 201 

§117. Definition of Integration. 118. Arbitrary Constant of 
Integration. 119. Integration of Entire Functions. 120. The 
Logarithmic Function. 121. Another Method of obtaining the 
Logarithmic Integral. 122. Exponential Functions. 123. The 
Elementary Forms of Integration. 

Chapter II. Integrals immediately reducible to the 
Elementary Forms 209 

§ 124. Integrals reducible to the Form / i^dy. 125. Appli- 
cation to the Case of a Falling Body. 126. Reduction to the Loga- 
rithmic Form. 127. Trigonometric Forms. 128. Integration of 

-T- ; — x and -= r. 129. Integrals of the Form / — — — ; — ■. 

a 2 -f- x 2 a 2 — x 2 ° J a-\-bx-\-cx 2 

130. Inverse Sines and Cosines as Integrals. 131. Two Forms of 
Integrals expressed by Circular Functions. 132. Integration of 

133. Integration of — . 134. Exponen- 



\/a 2 t x 2 ^a-\-bx ± ex' 

tial Forms. 

Chapter ILL Integration by Rational Transformations. . 222 
§ 135. Integration of - — — —dx, - — — =~r- and 



x n ' (a + bx) n a -\-bx ± cx r 

136. Reduction of Rational Fractions in general. 137. Integra- 
tion by Parts. 

Chapter IV. Integration of Irrational Algebraic Dif 

FERENTIALS 233 

§138. When Fractional Powers of the Independent Variable 
enter into the Expression. 139. Cases when the Given Differen- 
tial Contains an Irrational Quantity of the Form y'a -f- bx -f- ex 2 . 

dr 
140. Integration of dQ = — . 141. General Theory 

r yar 2 -j- br — 1 
of Irrational Binomial Differentials. 142. Special Cases when 
m -f- 1 = n, or m -\- 1 -f- np = — n. 143. Forms of Reduction 
of Irrational Binomials. 144. Formulae A and B, in which m is 
increased or diminished by n. 145. Formulas C and D, in which 
p is increased or diminished by 1. 146. Effect of the Formulae. 
147. Case of Failure in this Reduction. 



CONTENTS. IX 

PAGE 

Chaptek V. Integration of Transcendent Functions 246 

§ 148. Integration of / e mx cos nxdx and / e mx sin nxdx. 

149. Integration of sin m x cos ra xdx. 150. Special Cases of / sin m x 

cos n xdx. 151. Integration of . = . — 5 3—. 152. Integra- 

m 2 sin 2 x -\- 7v cos 2 # 

tion of f- . 153. Special Cases of the Last Two Forms. 

a -f- b cos y 

154. Integration of sin mx cos nxdx. 155. Integration by Devel- 
opment in Series. 

Chapter VI. Of Definite Integrals 255 

§156. Successive Increments of a Variable. 157. Differential 
of an Area. 158. The Formation of a Definite Integral. 159. Two 
Conceptions of a Definite Integral. 160. Differentiation of a 
Definite Integral with respect to its Limits. 161. Examples and 
Exercises in finding Definite Integrals. 162. Failure of the 
Method when the Function becomes Infinite. 163. Change of 
Variable in Definite Integrals. 164. Subdivision of a Definite In- 
tegral. 165. Definite Integrals through Integration by Parts. 

Chapter VII. Successive Integration 272 

§ 166. Differentiation under the Sign of Integration. 167. Ap- 
plication of the Principle to Definite Integrals. 168. Integration 
by means of Differentiating Known Integrals. 169. Application 
to a Special Case. 170. Double Integrals. 171. Value of a Func- 
tion of Two Variables obtained from its Second Derivative. 
172. Triple and Multiple Integrals. 173. Definite Double Inte- 
grals. 174. Definite Triple and Multiple Integrals. 175. Product of 

e-^dx. 

CO 

Chapter VIII. Rectification and Quadrature 285 

§ 177. The Rectification of Curves. 178. The Parabola. 179. The 
Ellipse. 180. The Cycloid. 181. The Archimedean Spiral. 
182. The Logarithmic Spiral. 183. The Quadrature of Plane 
Figures. 184. The Parabola. 185. The Circle and the Ellipse. 
186. The Hyperbola. 187. The Lemniscate. 188. The Cycloid. 

Chapter IX. The Cubature of Volumes 297 

§189. General Formulae. 190. The Sphere. 191. The Pyra- 
mid. 192. The Ellipsoid. 193. Volume of any Solid of Revolu- 
tion. 194. The Paraboloid of Revolution. 195. The Volume gen- 
erated by the Revolution of a Cycloid around its Base. 196. The 
Hyperboloid of Revolution of Two Nappes. 197. Ring-shaped 
Solids of Revolution. 198. Application to the Circular Ring. 
199. Quadrature of Surfaces of Revolution. 200. Examples of 
Surfaces of Revolution. 



PART I. 
THE DIFFERENTIAL CALCULUS. 



USE OF THE SYMBOL = 

The symbol = of identity as employed in this work indi- 
cates that the single letter on one side of it is nsed to repre- 
sent the expression or thing defined on the other side of it. 

When the single letter precedes the symbol =, the latter 
may commonly be read is put for, or is defined as. 

Wnen the single letter follows the symbol, the latter may 
be read ivliicli Jet us call. 

In each case the equality of the quantities on each side of 
= does not follow from anything that precedes, but is assumed 
at the moment. But having once made this assumption, any 
equations which may flow from it are expressed by the sign 
=, as usual. 



PART I. 
THE DIFFERENTIAL CALCULUS. 



CHAPTER I. 

OF VARIABLES AND FUNCTIONS. 



1. In the higher mathematics we conceive ourselves to be 
dealing with pairs of quantities so related that the value of 
one depends upon that of the other. For each value which 
we assign to one we conceive that there is a corresponding 
value of the other. 

For example, the time required to perform a journey is a 
function of the distance to be travelled, because, other things 
being equal, the time varies when the distance varies. 

We study the relation between two such quantities by as- 
signing values at pleasure to one, and ascertaining and com- 
paring the corresponding values of the other. 

The quantity to which we assign values at pleasure is called 
the independent variable. 

The quantity whose values depend upon those of the inde- 
pendent variable is called a function of that variable. 

Example I. If a train travels at the rate of 30 miles an 
hour, and if we ask how long it will take the train to trave] 
15 miles, 30 miles, 60 miles, 900 miles, etc., we shall have for 
the corresponding times, or functions of the distances, half an 
hour, one hour, two hours, thirty hours, etc. 



4 TEE DIFFERENTIAL CALCULUS. 

In thinking thus we consider the distance to be travelled as 
the independent variable, and the time as the function of the 
distance. 

Example II. If between the quantities x and y we have 
the equation 

y = 2ax*, 
we may suppose 

x = - 1, 0, + 1, -f 2, + 3, etc., 
and we shall then have 

y = 2a, 0, 2a, 8a, 18a, etc. 

Here x is taken as the independent variable, and y as the 
function of x. For each value we assign to x there is a corre- 
sponding value of y. 

When the relation between the two quantities is expressed 
by means of an equation between symbolic expressions, the 
one is called an analytic function of the other. 

An analytic function is said to be 

Explicit when the symbol which represents it stands 
alone on one side of the equation; 

Implicit when it does not so stand alone. 

Example. In the above equation y is an explicit function 
of x. But if we have the equation 

if -f xy = x% 

then for each value of x there will be a certain value of y, 
which will be found by solving the equation, considering y as 
the unknown quantity. Here y is still a function of x, be- 
cause to each value of x corresponds a certain value of y; bit 
because y does not stand alone on one side of the equation it 
is called an implicit function. 

Eemaek. The difference between explicit and implicit 
functions is merely one of form, arising from the different 
ways in which the relation may be expressed. Thus in the 
two forms 



VARIABLES AND FUNCTIONS. 5 

y = 2ax t > 

y — 2ax 2 = 0, 

y is the same function of x; but its form is explicit in the first 
and implicit in the second. 

An implicit function may be reduced to an explicit one by 
solving the equation, regarding the function as the unknown 
quantity. But as the solution may be either impracticable 
or too complicated for convenient use, it may be impossible to 
express the function otherwise than in an implicit form. 

2. Classification of Functions, When y is an rplicit 
function of x it is, by definition, equal to a symbolic ex\ 3ssion 
containing the symbol x. Hence we may call either y or the 
symbolic expression the function of x, the two being equiva- 
lent. Indeed any algebraic expression containing a symbol is, 
by definition, a function of the quantity represented by the 
symbol, because its value must depend upon that of the sym- 
bol. 

Every algebraic expression indicates that certain operations 
are to be performed upon the quantities represented by the 
symbols. These operations are: 

1. Addition and subtraction, included algebraically in one 
class. 

2. Multiplication, including involution. 

3. Division. 

4. Evolution, or the extraction of roots. 

A function which involves only these four operations is 
called algebraic. 

Functions are classified according to the operations which 
must be performed in order to obtain their values from the 
values of the independent variables upon which they depend. 

A rational function is one in which the only operations 
indicated upon or with the independent variable are those of 
addition, multiplication, or division. 



6 THE DIFFERENTIAL CALCULUS. 

An entire function is a rational one in which the only in- 
dicated operations are those of addition and multiplication. 
Examples. The expression 

a -f- hx + ex 2 -f- dx* 

is an entire function of x, as well as of a, b, c and d. 

The expression 

. m r. 



x x* -\-nx 
is a rational function of x, but not an entire function of x. 

An irrational function of a variable is one in which the 
extraction of some root of an expression containing that vari- 
able is indicated. 

Example. The expressions 



V a -f lx 9 (a -f mx % + nx*) 

are irrational functions of x. 

Functions which cannot be represented by any finite com- 
bination of the algebraic operations above enumerated are 
called transcendental. 

An exponential function is one in which the variable 
enters into an exponent. 

Example. The expressions 

(a + x) ny } x 2 * 

are entire functions of x when n and y are integers. But 
they are exponential functions of y. 

Other transcendental functions are: 

Trigonometric functions, the sine, cosine, etc. 

Logarithmic functions, which require the finding of a 
logarithm. 

Circular functions, which are the inverse of the trigo- 
nometric functions; for example, if 

y = a trigonometric function of x, sin x for instance, 
then x is a circular function of y, namely, the arc of which y 
is the sine. 



VARIABLES AND FUNCTIONS. 7 

3. Functional Notation, For brevity and generality we 
may represent any function of a variable by a single symbol 
having a mark to indicate the variable attached to it, in any 
form we may elect. Such a symbol is called a functional 
symbol or a symbol of operation. 

The most common functional symbols are 

F, f and 0; 

but any signs or mode of writing whatever may be used. 
Then, such expressions as 

Fix), f(x), 0(x), 
each mean 

" some symbolic expression containing x." 

The variable is enclosed in parentheses in order that the 
function may not be mistaken for the product of a quantity 
F, f or by x. 

Identical Functions. Functions which indicate identical 
operations upon two variables are considered as identical. 

Example. If we consider the expression 

a + by 

as a certain function of y, then 

a + bx 
is that same function of x, and 

a + b{x + y) 
is that same function of x -f- y. 

When the functional notation is applied, then: 
Identical functions are represented by the same functional 
symbols. 
Examples. If we put 

F(x) = a-\-bx, 

we shall have F(y) = a + by; 

F(f) =a+ bf; 
F(x* -f) = a + b(x> - tf). 



8 THE DIFFERENTIAL CALCULUS. 

In general, If we define afunctional symbol as representing 
a certain function of a variable, that same symbol applied to 
a second variable will represent the expression formed by sub- 
stituting the second variable for the first. 

In applying this rule any expression may be regarded as a 
variable to be substituted, as, in the last example, we used 
x 2 — y 2 as a variable to be substituted for x in the original 
expression. 

EXERCISES. 

1. If we put 

(p(x) = ax 2 , 

it is required to form and reduce the functions 

0(.y), 0(5), 0(a), <P{-x), 0(z 2 ), 0(|). 

2. Putting 

m = f±- x , 

it is required to form and reduce 

*+i>, J® '$ '(9- >©+.'©•' 

3. Putting 

it is required to form and reduce 

fix -a), f{x + a), /Q, /(|). 

4. If <p(x) = a 2 # + cz 2 , 
form and reduce the expressions 

0(z 2 ), 0(« 2 ), 4>{ax), <p{bx), 0(a + c), 0(a — c). 

5. Suppose 0(a) = ax 2 — a 2 #, and thence form 
<P(y), <P( Z )> <P(Py), 
<p(x + y), (p(x + «), <p{x — a), 
cp(x + ay), <p{x - ay), <p(x 2 ). 



VARIABLES AND FUNCTIONS. 9 

6. Suppose f{x) = # 2 , and thence form the values of 

f(l), f(x<), f{x% f(x% f(x% fix'). 

7. Let us put (p(m) = m{m — 1) (m — 2) (m — 3); thence 
find the values of 

0(6), 0(5), 0(4), 0(3), 0(2), 0(1), 0(0), 0(- 1), 0(- 2). 

8. Prove that if we put <fi{x) ~ a x , we shall have 

<P(x + y) = 0(z) x 0(^/); 0(sy) - [0(z)p s [#($]- 

4:, Functions of Several Variables* An algebraic expres- 
sion containing several quantities may be represented by any 
symbol having the letters which represent the quantities at- 
tached. 

Examples. We may put 

(p(x, y) = ax- by, 

the comma being inserted between x and y so that their 
product shall not be understood. We shall then have 

<p(m, n) === am — bn, 
<p(y, x)—ay — bx, 

the letters being simply interchanged; 

<p(x + y,x-y) = a(x + y)- b(x - y) 
= (a- b)x + (a + b)y; 
<p{a, b) = a* - Z> 2 ; 
<p(b, a) = ab — ba = 0; 
4>(a ~\- by ab) = a(a -f b) — ab*; 
0(a, a) = a 2 — ba; 
etc. etc. 

If we put 0(«, b, c) = 2a -{- Sb — be, we shall have 
<p(x, z, y) =2x + 3z — 5y; 
<p(z, y, x) = 2z + 3?/ — 5a; 
0(m, ???, — m) = 2m + 3m + 5??? = 10???; 
0(3, 8, 6) = 2-3 -f 3-8 - 5'6 = 0. 



10 THE DIFFERENTIAL CALCULUS. 

EXERCISES. 

Let us put <p(x, y) = 3x — fy; 

f(x, y) = ax + by, 
f{x, y, z)=ax + by — adz. 
Thence form the expressions: 

i. cp{y, x). 2. <p(a, I). 3- 0(3, 4). 

4 . 0(4, 3). 5. 0(10, 1). 6. f{a, b). 

7 . f(b, a). 8. f(y, x). 9 . /(7, - 3). 

IO - /fe — 'JP>- «• /(*> ^ 2/)* I2 - /(^ ^ 2 )- 

13. /(«, 6, c). 14. /(a% 6 2 , c 2 ). 15. f(-a,-b,-ab). 

Sometimes there is no need of any functional symbol 
except the parentheses. For example, the form (m, n) may 
be used to indicate any function of m and n. 

EXERCISES. 

T m(wi — 1) (m — 2) 
Let us put (m, n) = — ) -^j kt, 

r v ' w(w — 1) (w — 2) 

then find the values of — 

1. (3, 3). 2. (4, 3). 3. (5, 3). 

4. (6, 3). 5. (7, 3). 6. (8, 3). 

7. (2, - 1). 8. (3, - 2). 9. (4, - 2). 

5. Functions of Functions. By the definitions of the pre- 
ceding chapter, the expression 



/(#*)) 



will mean the expression obtained by substituting cp(x) for x 
mf{x). 

"We may here omit the larger parentheses and write f(p(x) 

instead of /( <fi(x) J . 

For example, using the notation of exercises 1 and 3 of 

§ 3, we shall have 

. , , x ax 2 — a x 7 — 1 



VARIABLES AND FUNCTIONS. 11 

For brevity we use the notation 

8 O*O = 0(0(aO). 

Continuing the same system, we have 

<p\x) = 4>[<t>\x)) =0' (#*)); 





etc. etc. etc. 


Examples. 


1. If 




<p(x) = ax 2 , 


en 


2 (z) = a(ax 2 y = flV; 




<f)\x) = a(a 3 xy = aV; 




etc. etc. etc. 



2. If 

f(x) = a — x, 

then f*( x ) = a — (a — x) = x; 

f\x) — a — f 2 (x) = a — x; 
and, in general, 

f"-\x) =f n (x). 
Remakk. The functional nomenclature may be simplified 
to any extent. 

1. The parentheses are quite unnecessary when there is no 
danger of mistaking the form for a product. 

2. When it is once known what the variables are, we may 
write the functional symbol without them. Thus the symbol 
may be taken to mean <px or <p(x). 

6. Product of the First n Numbers. The symbol n\, called 
factorial n, is used to express the product of the first n num- 
bers, 

1-2-3-...W. 

Thus, 1!=1; 

2! = 1-2 = 2; 
3! = 1-2-3 = 6; 
4! = 1-2-3-4 = 24; 
etc. etc. 



12 THE DIFFERENTIAL CALCULUS. 

It will be seen that 

2! = 2-1!; 
3! = 3-2!; 
and, in general, n\ = n ' (n — 1)!, 

whatever number n may represent. 

EXERCISES. 

Compute the values of — 

i. 5! 2. 6! 3. 8! 

7! 8! 

4 ' 3! 4! 5 ' 3! 5! 

6. Prove the equation 2 • 4 • 6 ■ 8 • . . . %n — 2 n n ! 

7. Prove that, when n is even, 

n } __ n(n — 2) (n —4)... 4- 2 

2" ~~ re 

2 2 

7 . Binomial Coefficients. The binomial coefficient 

w(w — 1) (n — 2) to s terms 

1-2-3-...S 

is expressed in the abbreviated form 



fi). 



the parentheses being used to distinguish the expression from 
n 



the fraction -. 

s 






EXAMPLES. 




eK-* 




/7\7-6-5-4-3_ 
\5/ 1-2-3-4-5 




lrJ=r = M - 




^\ n(n — l)(n — 2) 



1-2-3 



VARIABLES AND FUNCTIONS. 



13 



EXERCISES. 



Prove the formulae: 



I. \tt = 



2 J U - 2/ 



V2/ 2! 3! 



+ 1 



2. 



4- - = 



w: 



)+©=( 



1 
» + l 



a! (w — 5)! 

s + © - m 

+ fi) - m 



8. Graphic Representation of Functions. The methods of 
Analytic Geometry enable ns to represent functions to the eye 
by means of curves. The common way of doing this is to 
represent the independent variable by the abscissa of a point, 
and the corresponding value of the function by its ordinate. 
Let x l9 # 2 , x 3 , etc., be 
different values of the in- y 

dependent variable, and 

y*> y*> y*> etc -> the cor - 

responding values of the 
function. We lay off 
upon the axis of abscis- To 
sas the lengths OX lf '-- 

0X 2 , 0X % , etc., equal 
to x l9 x 2 , ic 3 , etc., and terminating at the points X lf X 9 , X s , 
etc. At each of these points we erect a perpendicular to rep- 
resent the corresponding value of y. The ends, P it P 9 , P„ 
of these perpendiculars will generally terminate on a curve 
line, the form of which shows the nature of the function. 

It must be clearly seen and remembered that it is not the 
curve itself which represents the values of the function, but 
the ordinates of the curve. 







P2 Pa 




Pl- 






























K 




V\ 


& 


Vz 














I / JXl X2 X3 















-X 



Fig. 1. 



14 



THE DIFFERENTIAL CALCULUS. 



Fig. 2. 



9. Contimiity and Discontinuity of Functions. Let us 
consider the graphic representation of a function in the most 
general way. We measure off a series of values, 0X l9 0X„ 
0X % , etc., of the independent variable, and at the points X l9 
X if X s , etc., we erect ordinates. 
In order that the variable ordinate 
may actually be a function of x it is 
sufficient if, for every value of the 
abscissa, there is a corresponding 
value of the ordinate. 

Now we might conceive of such 
a function that there should be no ~i ' i ' ' i i ' ' ' i ' i ' ' '^ 
relation between the different val- 
ues of the ordinates, but that every 
separate point should have its own 
separate ordinate, as shown in 
Fig. 2. ^ If this remained true how 
numerous soever we made the ordi- 
nates, then the ends of the latter would not terminate in any 
curve at all, but would be scattered over the plane. Such 
a function would be called discontinuous at every point. 

Such, however, is not the kind of functions commonly 
considered in mathematics. The functions with which we 
are now concerned are such that, however irregular they may 
appear when the values of x are widely separated, the ends 
of the ordinates will terminate in a curve when we bring 
those values close enough together. 

If a function is such that when the point representing the 
independent variable moves continuously from X 1 to X 2 (Fig. 
1) the end of the ordinate describes an unbroken curve, then 
we call the function continuous between the values x l and 
# a of the independent variable. 

If the curve remains unbroken how far soever we suppose 
x to increase, positively or negatively, we call the function 
continuous for all values of the independent variable. 



VARIABLES AND FUNCTIONS. 



15 



But if there is a value a of x for which there is a break of 
any kind in the curve, we call the function discontinuous for 
the value a of the independent variable. 

Let us, for example, consider the function 



y 



5(a — x)' 



Let us measure off on the axis of abscissas the length OX 
= a . Then as we make our varying ordinate approach X 
from the left it will increase positively without limit, and the 
curve will extend upwards to infinity; if we approach X from 
the right-hand side, the ordinate will be negative and the 
curve will go downwards to infinity. Thus the curve will not 
form a continuous branch from the one side to the other. 
Thus the above function is discontinuous for the value a of x. 



i 

Fig. 3. 

10. Many-valued Functions. In all that precedes, we 
have spoken as if to each value of the independent variable 
corresponded only one value of the function. But it may 



16 



THE DIFFERENTIAL CALCULUS. 



happen that there are several such values. For example, if y 
is an implicit function of x represented by the equation 

y* + mxy* -j- nx *V + P x * *= ®> 

then we know, by the theory of equations, that there will be 
three values of y for each value assigned to the variable x. 

Defi According as a function admits of one, two or n 
values, it is called one-valued, two-valued or w-valued. 

Infinitely -valued Functions. It may happen that to each 
value of the variable there are an infinity of different values 
of the function. A case of this is the function sin (_ x) x, or 
the arc of which x is the sine. This arc may be either the 
smallest arc which has x for its sine, or this smallest arc in- 
creased by any entire number of circumferences. 

Take, for example, the arc whose sine shall 
be+|. 

The two smallest arcs will be 

30° = \n and 150° == \n. 

But if we take the function in its most gen- 
eral sense it may have any of the values 

(2+i)*r; (4 + i)*; (6 + j>r, etc., 
or (2+f)7r; (4 + |)7r; (6 + *)*r, etc. 

When we represent an ^-valued function 
graphically, there will be n values to each ordi- 
nate. Hence each ordinate will cut the curve 
in n points, real or imaginary. 

The figure in the margin represents the infi- 
nitely-valued function 

x 




y = a sin 



(-D 



When — a < x < + a, any ordinate will cut 
the curve in an infinity of points. 



o 



LIMITS AND INFINITESIMALS. 17 



CHAPTER II. 
OF LIMITS AND INFINITESIMALS. 

11. Limits. The method of limits is an indirect method 
of arriving at the values of certain quantities which do not 
admit of direct determination. The method rests upon the 
following axioms and definition: 

Axiom I. Any quantity, however small, may be multiplied 
by so great a number as to exceed any other quantity of the 
same kind, however great, to which a fixed value is assigned. 

Axiom II. Conversely, any quantity, however great, may 
be divided into so many parts that each part shall be less than 
any other quantity of the same kind, however small, to which 
a fixed value is assigned. 

Axiom III. Any quantity may be divided into any num- 
ber of parts ; or multiplied any number of times. 

Def. The limit of a variable quantity X is a quantity L, 
which we conceive Xto approach in such a way that the dif- 
ference L — X becomes less than any quantity we can name, 
but which we do not conceive X to reach. 

Example. If we have a variable quantity X and a con- 
stant quantity L, and if X, in varying according to any mathe- 
matical law, takes the successive values 

L ± 0.1, 
L ± 0.01, 
L ± 0.001, 
L ± 0.0001, 

and so on indefinitely, without becoming equal to L, then we 
say that L is the limit of x. 



18 TEE DIFFERENTIAL CALCULUS. 

12. Infinites and Infinitesimals. Definitions. 

1. An infinite quantity is one considered as becoming 
greater than any quantity which we can name. 

2. An infinitesimal quantity is one considered in the 
act of becoming less than any quantity which we can name; 
that is, in the act of approaching zero as a limit. 

3. A finite quantity is one which is neither infinite nor in- 
finitesimal.* 

Examples. If of a quantity x we either suppose or prove 

x > 10, 
x > 100, 
x > 100000, 

and so on without end, then x is called an infinite quantity. 
If of a quantity h we either suppose or prove 

h < 0.1, 
h < 0.001, 
h < 0.00001, 

and so on without end, then h is an infinitesimal quantity. 

The preceding conceptions of limits, infinites and infinitesi- 
mals are applied in the following ways: Let us have an inde- 
pendent variable x, and a function of that variable which we 
call y. 

Now, in order to apply the method of limits, we may make 
three suppositions respecting the value of x, namely: 

1. That x approaches some finite limit. 

2. That x increases without limit (i.e., is infinite). 

3. That x diminishes without limit (i.e., is infinitesimal). 
In each of these cases the result may be that y approaches 

a finite limit, or is infinite, or is infinitesimal. 

* Strictly speaking, the words infinite and infinitesimal are both adjec- 
tives qualifying a quantity. But the second has lately been used also as 
a noun, and we shall therefore use the word infinite as a noun meaning 
infinite quantity. 



LIMITS AND INFINITESIMALS. 19 

For example, let us have 

x 4- a 

y — — ! — • 

9 x — a 
Then— 

When x approaches the limit a, y becomes infinite. 

When x becomes infinite, y approaches the limit + 1- 

When x becomes infinitesimal, y approaches the limit — 1. 

The symbol =, followed by that of zero or a finite quantity, 
means " approaches the limit." The symbols ^oo mean 
" increases without limit" or "becomes infinite." Hence 
the three last statements may be expressed symbolically, as 
follows: 

x -f a . 



When x = a, then 



x — a 



When x = oo , then = -4- 1 ; 

x — a 

etc. etc. 

The same statements are more commonly expressed thus: 

,. x + a . . x 

hm. — ! — (x - a) = oo : 

hm. ^^ (* = <*>) = +1; 

li m . ^±_? ( X ± o) = _ i. 
x — a x ' 

13. Properties of Infinite and Infinitesimal Quantities. 

Theokem I. The product of an infinitesimal by any finite 
factor, however great, is an infinitesimal. 

Proof. Let h be the infinitesimal, and n the finite factor 
by which it is multiplied. I say how great soever n may be, 
nh is also an infinitesimal. For, if nh does not become less 
than any quantity we can name, let a be a quantity less than 
which it does not become. Then if we take, as we may, 

h < jj, (Axiom III.) 

we shall have nh < a. 



20 THE DIFFERENTIAL CALCULUS. 

That is, nh is less than a and not less than a, which is 
absurd. 

Hence nh becomes less than any quantity we can name, 
and is therefore infinitesimal, by definition. 

Theorem II. The quotient of an infinite quantity by any 
finite divisor, however great, is infinite. 

Proof. Let X be the infinite quantity, and n the finite 
divisor. If X -s- n does not increase beyond every limit, let 
K be some quantity which it cannot exceed. Then by taking 

X>nE, (Ax. III.) 

X 

we shall have — > K\ 

n 

that is, — greater than the quantity which it cannot exceed, 

which is absurd. 

Hence X ~ n increases beyond every limit we can name 
when X does, and is therefore infinite when X is infinite. 

Theorem III. The product of any finite quantity, how- 
ever small, by an infinite multiplier, is infinite. 

This follows at once from Axiom I., since by increasing the 
multiplier we may make the product greater than any quan- 
tity we can name. 

Theorem IV. The quotient of any finite quantity, how- 
ever great, by an infinite divisor is infinitesimal. 

This follows at once from Axiom II., since by increasing 
the divisor the quotient may be made less than any finite 
quantity. 

Theorem V. The reciprocal of an infinitesimal is an in- 
finite, and vice versa. 

Let h be an infinitesimal. If 7- is not infinite, there must 
be some quantity which we can name which j does not ex- 



LIMITS AND INFINITESIMALS. 21 

ceed. Let K be that quantity. Because h is infinitesimal, 
we may have 

h <i> 

which gives =-> K; 

that is, j greater than a quantity it can never exceed, which 

is absurd. 

The converse theorem may be proved in the same way. 

14. Orders of Infinitesimals. Def. If the ratio of one 
infinitesimal to another approaches a finite limit, they are 
called infinitesimals of the same order. 

If the ratio is itself infinitesimal, the lesser infinitesimal is 
said to be of higher order than the other. 

Theorem VI. If we have a series proceeding according 
to the powers of h, 

A+BU+ Oh 1 + Dh 3 + etc., 

in which the coefficients A, B, C, are all finite, then, if h be- 
comes infinitesimal, each term after the first is an infinitesi- 
mal of higher order than the term preceding. 

Proof. The ratio of two consecutive terms, the third and 
fourth for example, is 

m _D 
Cti ~ G 

By hypothesis, (7 and D are both finite; hence -^ is finite; 

hence when h approaches the limit zero, -^h becomes an in- 

finitesimal (§13, Th. I.). Thus, by definition, the term Dh* 
is an infinitesimal of higher order than Olr. 

Def. The orders of infinitesimals are numbered by taking- 
some one infinitesimal as a base and calling it an infinitesi- 
mal of the first order* Then, an infinitesimal whose ratio to 



22 THE DIFFERENTIAL CALCULUS. 

the nth power of the base approaches a finite limit is called 
an infinitesimal of the nth order. 

Example. If h be taken as the base, the term 

Bh is of the first order ■ . ■ Bh : h = the finite quantity B; 
CW " " second" • .• CV : h % = " « C; 

Eh n " " nth " '.'Eh n ;h n = " " E. 

Cor. 1. Since when n = we have Bh n = Bh° = B for 
all values of h 3 it follows that an infinitesimal of the order 
zero is the same as a finite quantity. 

Cor. 2. It may be shown in the same way that the product 
of any two infinitesimals of the first order is an infinitesimal 
of the second order. 

15, Orders of Infinites. If the ratio of two infinite 
quantities approaches a finite limit, they are called infinites 
of the same order. 

If the ratio increases without limit, the greater term of the 
ratio is called an infinite of higher order than the other. 

Theokem VII. In a series of terms arranged according 
to the powers of x, 

A -f Bx + Cx* -f Dx 3 + etc., 

if A, B, C, etc., are all finite, then, when x becomes infinite, 
each term after the first is an infinite of higher order than the 
term preceding. 

For, the ratio of two consecutive terms is of the form -^x, 

which becomes infinite with x (Th. III.). 

Def. Orders of infinity are numbered by taking some one 
infinite as a base, and calling it an infinite of the first order. 

Then, an infinite whose ratio to the wth power of the base 
approaches a finite limit is called an infinite of the nth. order. 

Thus, taking x as the standard, when it becomes infinite 
we call Bx infinite of the first order, Cx* of the second order, 
etc. 



LIMITS AND INFINITESIMALS. 23 



NOTE ON THE PRECEDING CHAPTERS. 

In beginning the Calculus, conceptions are presented to the student 
which seem beyond his grasp, and methods which seem to lack rigor. 
Really, however, the fundamental principle of these methods is as old 
as Euclid, and is met with in all works on elementary geometry which 
treat of the area of the circle. The simplest form in which the princi- 
ple appears is seen in the following case. 

Let us have to compare two quantities A and B, in order to determine 
whether they are equal. If they are not equal, then they must differ by 
some quantity. If, now, taking any arbitrary quantity h, we can prove 
that 

A-B<h 

without making any supposition respecting the value of h, this will show 
that A and B are rigorously equal. For if they differed by the quantity 
a, then when h was less than a the above inequality would not hold 
true. But as we have been supposed to prove it for all values of h, it 
must be true when h is less than a. In this case h might be considered 
an infinitesimal, although in the Elements of Euclid it is represented on 
the page of the book by a figure nearly an inch square. 

Infinitesimal quantities were formerly called infinitely small. When 
they were introduced by Leibnitz many able mathematicians were unable 
to accept them. Bishop Berkeley wrote several essays against them, in 
one of which he suggested that they might be called the ghosts of departed 
quantities. The following propositions are presented in the hope that 
they may save the student unnecessary efforts of thought in the study of 
this subject. 

• Firstly, there is no need that a quantity should be considered as ab- 
solutely infinite. A mathematical magnitude, considered as a quantity, 
must in its very nature have boundaries, because mathematics is con- 
cerned with the relation between magnitudes as greater or less, and 
we can compare two magnitudes as greater or less only b} r comparing 
their boundaries. An absolutely infinite magnitude, having no boun- 
daries to compare, cannot be compared with anj^thing. 

Secondly, it is equally unnecessary to suppose the existence, either in 
nature or in thought, of quantities which are absolutely smaller than 
any finite quantity whatever. 



24 THE DIFFERENTIAL CALCULUS. 

But however small a quantity may be, there may always be another 
still smaller in any ratio. Hence, although it is perfectly true that no 
quantity can be otherwise than finite, yet it is equally true that a quantity 
may be less or greater than any fixed quantity we may name. 

Both infinite and infinitesimal quantities are therefore essentially in- 
definite, because by considering them in the act of increasing beyond, or 
decreasing below, every assignable value, we do away with the very pos- 
sibility of assigning values to them. They are used only as auxiliaries 
to lead us to a knowledge of finite quantities, and their magnitudes are 
never themselves the object of investigation. 

The essentially indefinite nature of infinites and infinitesimals may be 
illustrated as follows: 

If we have an equation of the form 

x —9* 

then for every pair of finite values we assign to a and b there will be a 
definite value of x. 

But if we suppose A and B to be infinite, and at the same time inde- 
pendent of each other, there will be no definite value to x. Considering 
both terms as absolutely infinite, they will have no bounds, and there- 
fore cannot be compared in value. Considered as increasing without 
limit, one may be any number of times greater than the other, and thus 
the fraction may have any value we choose to assign it. Seeking for 
the value of such a fraction is like trying to answer the old question 
concerning the effect of an irresistible force acting upon an immovable 
obstacle. 



DIFFERENTIALS AND DERIVATIVES. 25 



CHAPTER III. 

OF DIFFERENTIALS AND DERIVATIVES. 

16. Be/ An increment of a variable is the difference 
between two values of that variable. 

An equivalent definition is: An increment is a quantity 
added to one value of a variable in order to obtain another 
value. 

Notation. An increment is expressed by the symbol A 
written before the symbol of the variable. 

Example. If we have the different variables 

x, y, u, 
and the increments Ax, Ay, Au, 
other values of the variables will be 

x -|- Ax, y -f- Ay, u -f- Au. 

Here A is not a factor multiplying x, but a symbol meaning 
"increment of/' or, in familiar language, "a little piece of." 

In considering the respective increments of an independent 
variable, and of its function, the following five quantities 
come into play and are each to be clearly conceived. 

1. A value o/ the independent variable, which we may take 
at pleasure. 

2. The cor responding value of the function, which will be 
fixed by that of the independent variable. 

3. An increment of the independent variable, also taken at 
pleasure. 

4. The corresponding increment of the function, deter- 
mined by that of the independent variable. 

5. The ratio of these increments. 



26 



THE DIFFERENTIAL CALCULUS 



To represent these quantities, let the relation between the 
variable x and the function y be expressed by a curve. Let 
OX be one value of x, and OX' another. Let XP and X'P' 




Fig. 5. 

be the corresponding values of y, leading to the points P and 
P' of the curve. We shall then have — 

1. OX — x, a value of the independent variable. 

2. XP = y, the corresponding value of the function. 

3. XX' = Ax, an arbitrary increment of x. 

4. RP' = Ay, the corresponding increment of y. 

Ay 

5. Then, by Plane Trigonometry, the quotient -~ will be 

the tangent of the angle PQX; that is, the tangent of the 
angle which the secant PP' makes with the axis of abscissas. 
Thus we have geometrical representations of the five fun- 
damental quantities under consideration. 

17. First Idea of Differentials and Derivatives. Let us 
take, for illustration, the function 

y = n x\ (1) 

Giving to x the increment Ax, the new value of y will be 

n{x + Ax)\ 

Hence y -f- Ay = n(x -f- Ax)* = nx 1 -\- 2nxAx -j- nAx*. (2) 



DIFFERENTIALS AND DERIVATIVES. 27 

Subtracting (1) from (2), we have, for the increment of y, 

Ay = n(2x + Ax) Ax, (3) 

Because, when Ax becomes infinitesimal, 
lim. (2x -\- Ax) — 2x, 

we have, for the ratio of the increments, 

Ay 

— ^ = 2nx -f nAx, (4) 

and, when Ax becomes infinitesimal, 

lim. ^ = 2nx. (5) 

Def. The differential of a quantity is its infinitesimal 
increment; that is, its increment considered in the act of ap- 
proaching zero as its limit, or of becoming smaller than any 
quantity we can name. 

Notation. The differential of a quantity is indicated by 
the symbol d written before the symbol of the quantity. 

For example, the expressions 

dx, da, d(x -f- y), 

mean any infinitesimal increments of x, u, (x -J- y), respect- 
ively. 

Thus the substitution of d for A in the notation of incre- 
ments indicates that the increment represented by A is sup- 
posed to be infinitesimal, and that we are to consider the limit 
toward which some quantity arising from the increment then 
approaches. 

Using this notation, the equation (5) may be written 

-/- = 2nx. 
dx 

We also express this value of the limiting ratio in the form 

dy = 2?ixdx; 

meaning thereby that the ratio of the two members of this 
equation has unity as its limit. This is evident from Eq. (3) 



28 



THE DIFFERENTIAL CALCULUS. 



Def. If y is a function of x, the ratio -~- of the differential 

of y to that of x is called the derivative of the function, or 
the derived function. 

18. Illustrations. As the logic of infinitesimals offers 
great difficulties to the beginner, some illustrations of the 
subject may be of value to him. 

Consider the following proposition: 

The error introduced by neglecting all the powers of an in- 
crement above the first may be made as small as tve please by 
diminishing the increment. 

Let us suppose n = 2 in the equation (1). We then have 
the equations 

y 



2x 2 ; 
Ay — 4:xJx 
Ay 
Ax 



= ±x 



-f 2 Ax'; 
2Ax. 



(a) 



The ratio of the two terms of the second member is 
2Ax Ax 

-JT' or Tx 



Let us now neglect this quantity and write the erroneous 
equation 

Ay 



Ax 



= Ax. 



If, now, 

we 
suppose 



Ax < m 



<J Ax < - 
Ax < 



10000' 

x 



the equation 
(b) will still be { 
true within 



200 
1 
20000 
1 



(?) 

part; 

part; 
part; 



1000000 J L 2000000 

etc., etc. 

So long as we assign any definite value to Ax, it is clear 
that there will be some error in neglecting Ax. But there is 
no error in the equations 

dy = 4:xdx and ~ = £x, 
ax 



DIFFERENTIALS AND DERIVATIVES. 29 

provided that we interpret them as expressing the limit which 

Ay 

-p approaches as Ax approaches the limit zero, and interpret 

all our results accordingly. 

19. Illustration oy Velocities. Let us consider what is 
meant by the familiar idea of a train which may be contin- 
ually changing its speed passing a certain point with a certain 
speed. To fix the ideas, suppose the train has just started 
and is every moment accelerating its speed in such manner 
that the total number of feet it has advanced is equal to the 
square of the number of seconds since it started. Put 

6 = the distance travelled expressed in feet; 
t = the time expressed in seconds. 

We shall then have 3 = t\ 

and for the distances travelled: 
Number of seconds, 0; 
Distance travelled, 0; 
Distance in each second, 

ill 3 , 5 , 7 

I s 2 s 3 s ' 4* 5* 

Fig. 6. 

Let this line represent the space travelled the first five 
seconds from the starting time, and let us inquire with what 
velocity the train passed the point B at the end of 4 9 . 

Since distance travelled = velocity x time, the mean ve- 
locity is found by dividing the space by the time required to 
pass over that space. Now, the train had travelled 

16 feet in the time 4 seconds, 
and (4 + Aty feet in (4 + At) seconds, 

or 16 + 8 At + Af feet in (4 -f At) seconds. 

Subtracting 16 feet and 4 seconds, wc sec that in the time 
At after the end of the 4 seconds the train went SAt -f- At* 
= As feet. Hence its mean velocity from 4 s to 4 s -f- At is 



i; 


2; 3; 


4; 


5; 


etc.; 


i; 


4; 9; 


16; 


25; 


etc.; 


i; 


3; 5; 


?; 


9; 


11; etc 


B 

1 


9 




11 





30 



THE DIFFERENTIAL CALCULUS. 



As 
At 



(8 -f- At) feet per second* 



Now it is clear that, since the train was Continually accel- 
erated how small soever we take At, the mean velocity during 
this interval will exceed that with which it passed B. But 
it is also clear that by supposing At to approach the limit zero, 
we shall approach the required velocity as our limit. Hence 
the velocity with which B was passed is rigorously 

ds 

Tt 



8 feet per second. 




Fig. 7. 

20. Geometrical Illustration. If, in the figure, we sup- 
pose the point P' to approach P as its limit, the increments 
Ax and Ay will approach the limit zero, and the secant P'P 
will approach the tangent at the point P as its limit. We 
have already shown that 

Aij 

—^ = tangent of angle made by secant with axis of abscissas. 

Passing to the limit, we have the rigorous proposition 

~f- = tangent of angle which the tangent at the point P 
makes with the axis of abscissas. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 31 



CHAPTER IV. 
DIFFERENTIATION OF EXPLICIT FUNCTIONS. 

2 1 . Def. The process of finding the differentia] and the 
derivative of a function is called differentiation. 

As exemplified in §§ 16, 17, it may be generalized as fol- 
lows: We have given 

(1) An independent variable = x. 

(2) A function of that variable = <p(x). 

(3) We assign to x an increment = Ax; whereby (p(x) is 
changed into <p(x -f- Ax). 

(4) We thus have <p(x -\- Ax) — (p(x) as the increment of 
(f)(x). We may put 

A<p{x) = <p(x + Ax) — <p(x). 

(5) We then form the ratio 

A0(x) 



Ax 



(a) 



and seek its limit when Ax becomes infinitesimal. Using the 
notation of the last chapter, we have 

, ' = lim. y ' - (Ax = 0), 
dx Ax v }i 

which is the derivative of cp(x). 

In order to find the ratio (a), it is necessary to develop 
(p(x -j- dx) in powers of Ax to at least the first power of Ax. 
Let this development be 

<f>{x + Ax) = X + X x Ax + X n Ax* + (1) 

In the second member of this equation X , X,, etc., will be 
functions of x; and it is evident that X can be nothing but 



32 TEE DIFFERENTIAL CALCULUS. 

cp(x) itself, because it is the value of <f>(x -f Ax) when Ax — 0. 
Thus we have 

A<p(x) = <p(x + Ax) - <p(x) = {X x + X,Ax) Ax -{- . . . ; 

Passing to the limit, 

clcp(x) — X x dx\ 
d^x) = 
dx x v 

Thus, by comparing with (1), we have the following: 

Theorem I. The derivative of a function is the coefficient 
of the first poiver of the increment of the independent variable 
when the function is developed in powers of that increment. 

If we have to differentiate a function of several 'variable 
quantities, x, ?/, z, etc., we assign an increment to each vari- 
able, and develop the function in powers and products of the 
increments. 

Subtracting the original function, the remainder will be its 
increment. 

The terms of highest order in this increment, considered 
as infinitesimals, are then called the differential of the 
function. 

The following are the special cases by combining which all 
derivatives of rational functions may be found. 

22. Differentials of Sums. Let x, y, z, u, etc., be any 
variables or functions whatever. Their sum will be 

x + y ~\~ z + u + e "t°' 

Assigning to each an increment, x will become x -f- Ax, y 
will become y -f- Ay, etc. Hence the sum will become 

x -f Ax -f y -f- Ay -f z -f Az -\- u -f- An -j- etc. 

Subtracting the original expression, we find the increment of 
the sum to be 

Ax -\- Ay -f Az -f Au -f etc. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 33 

Hence, when the increments become infinitesimal, 
d(x -f- y + # + w + etc.) = dx -{- dy -{- dz -\- du -{- etc., (3) 

or, in words: 

Theorem II. The differential of the sum of any number 
of variables is equal to the sum of their differentials. 

In this theorem the quantities x, y, z, u, etc., maybe either 
independent variables, or functions of one or more variables. 

23. Differential of a Multiple. Let it be required to find 
the differential of 

ax, 
a being a constant. 

Giving x the increment Ax, the expression will become 

a(x -f- Ax). 
Then, proceeding as before, we find 

d(ax) = adx. (4) 

24. Theorem III. The differ 'ential of any constant is 
zero. 

For, by definition, a constant is a quantity which we sup- 
pose invariable, and to which we cannot, therefore, assign any 
increment. 

We therefore have, from Theorem I. when x is a variable 
and a is a constant, 

d(x -J- a) = dx + == dx, 

or, in words: 

Theorem IV. The differential of the sum of a constant 
and a variable is equal to the differential of the variable alone. 

Remark. It will be readily seen that the conclusions of 
§§ 22-24 are equally true whether we suppose the increments 
to be finite or infinitesimal. This is no longer the case when 
powers or products of some finite increments enter into the 
expression for other finite increments. 



34 THE DIFFERENTIAL CALCULUS. 



EXERCISES. 

It is required, by combining the preceding processes, to 
form the differentials of the following expressions, supposing 
a, b and c to be constants, and all the other literal symbols 
to be variables. 

I. U — V. 2. %U — V. 

3. v -j- x -f- c. 4. ax -j- by. 

5. cfx + Vy -\-c. 6. Sx -\- ±ay + b. 

7. 4zax -j- 5bx — y. 8. 6bx — abc. 

9. 3x — a -J- ab. 10. «&& — «Ztf. 

11. c(2x -j- «). I2 - a(bx + ac). 

13. «c(#w -j- a#). 14. bc(2ax — Sby). 

15. x — y — z. 16. — a£ — by — cz. 

17. — a(bx — cy). 18. — b(2ax — Scv). 

x x -\- y — z 

19. -. 20. v • 

« 

21. (a + J + c) (s + tf + 3u - %). 

25. Differentials of Products and Powers. Take first 
the product of two variables, which we shall call w and v. 

Then 

Product = uv. 

Assigning the increments An and Av, the product becomes 

{u -J- Z/?^) (0 + ^ v ) — nv + v ^ w + u ^ v + AuAv. 

Subtracting the original function, uv, we find 

A(uv) — vAu -f- (u + Jw) Jy. 

Supposing the increments to become infinitesimals, the co- 
efficient of Av in the second member will approach u as its 
limit. Hence, passing to the limit (§ 14), 

d(uv) = vdu -f- udv* 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 35 

To extend the result to any number of factors, let P be the 
product of all the factors but one, and let the remaining fac- 
tor be x, so that we have 

Product = Px. 
By what precedes, we have 

d(Px) = xdP + Pdx. 

Supposing P to be a product of the two variables u and v, 
this result gives 

d(uvx) = xd(vu) -j- uvdx = vxdu -j- uxdv + uvdx. (a) 
If we add a fourth factor, y, we shall have 
d(uvxy) = yd(uvx) -j- uvxdy. 

If we substitute for d(uvx) its value (a), we see that we 
pass from the one case to the other by (1) multiplying all the 
terms of the first case by the common factor y; (2) adding 
the product of dy into all the other factors. 

We are thus led to the conclusion: 

Theoeem V. The differential of the product of any num- 
ber of variables is equal to the sum of the products formed by 
replacing each variable by its differential. 

Corollary. If the n factors are all equal, their product will 
become the nth. power of the variable, aud the n differentials 
will all become equal. Hence, when n is an integer, we have 
the general formula 

d(x n ) = x n ~ x dx + x n ~ 1 dx -f etc., to n terms, 
or d(x n ) = nx n ~ l dx. 

By combining the preceding processes we may form the 
differential of any entire function of any number of variables. 

Examples. 

i. d(ax + bxy -f- cxyz) 

= d(ax) + d(bxy) + d{cxyz) (Th. II. 22) 

= adx + bd(xy) + cd(xyz) (Th. III. 23) 

= adx -f b{ydx -f xdy) -f c(yzdx + xzdy + xydz) 
= (a + % -f- cyz)dx + (J» + cxz)dy + cxydz. 



36 THE DIFFERENTIAL CALCULUS. 

2. d(ax 3 + b) = rf(az 3 ) (Th. IV.) 

= ^(tf 3 ) (§ 23) 

= Batfdx. (Th. V., Cor.) 

3. d(ax 5 y n ) = ad{x z y n ) (§ 23) 

= %^(«') + *W)] (Th. V.) 

= 3ay n x\lx + nax 5 y n -hly. (Th. V., Cor.) 

4. tf(a + z a ) M = *(a + ^) n - 1 d(a + z 2 ) = 2»(a + ^) U -Hcdx. 



EXERCISES. 

Form the differentials of the following expressions, suppos- 
ing the letters of the alphabet from a to n to represent con- 
stants: 

i. a -f- bx 2 -f- c£ 4 . .4ws. (2&e -\-4=cx*)dx. 



2. 


B+Cy + Dy\ 


3. 


axy. 


4- 


fcft^g. 


5- 


a(x -f- yz). 


6. 


«(# 2 -f- buv). 


7. 


axy -f- &wv. 


8. 


h(x*y + xy*). 


?• 


ax 2 y 3 . 


IO. 


bx*y n . 


11. 


abx*y n -f ku m v n . 


12. 


z(mx -j- wy). 


13. 


(r + ff)(t + g). 


14. 


%(# — a; 2 ). 


15. 


ax 2 — byz. 


l6. 


(a + x) (b - y). 


17. 


(a + a?)(b-y>). 


l8. 


{a — x) (a —x" 2 ). 


19. 


x(a -f z) (£ - z 2 ), 


20. 


(A+Bx+Cx*)(y + z). 


2 2 


ay 


21. 


(A+Bf+Cy*)(ay+bx) 




• « * 


2 3- 


(a -J- bit*) (ex" 2 — «y*). 


24. 


x — wv 


25. 


(a- x)(b-x')(c- x z ). 


*■ -f . 


a 




x — uv, , N 


27. 


x\x* + y{a, - x)\ 


26. 

29. 


(ay' - bx*) (x - y). 


28. 


( x * L ?A 


30. 


C- -«(?+!> 


3 1 - 

3 2 - 


(rt+a;) 3 . 
n(a + a) 3 . 


33. 


(a + ay) 2 . 


34- 


(aa + fy) 2 . 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 37 

26 Differential of a Quotient of Two Variables. Let the 
variables be x and y, and let q be their quotient. Then 

x 

and qy — x. 

Differentiating, we have 

ydq + qdy = dx. 
Solving so as to find the value of dq, 

_ dx — qdy _ ydx — xdy 

Cq ~~ y if 

Hence: 

Theorem VI. The differential of a fraction is equal to 
the denominator into the differential of the numerator, minus 
the numerator into the differential of the denominator, divided 
by the square of the denominator. 

Remark. If the numerator is a constant, its differential 
vanishes, and we have the general formula 

d— = Jlx. 

x x 

EXERCISES. 

Form the differentials of the following expressions: 



a -\- x 



2. 



3- 



a + y a + y 

a — x x 2 

a - y if 



a a 

s ' * v W+y)" 

a -f ox m -j- nx* 
a -{-by 

x + y mx % -f ni? 

9' -• io. — -„ — a . 

x — y mx — ny 



38 THE DIFFERENTIAL CALCULUS. 

a x -\- yz 



a-\- bx -f- ex 
m + wy 
I3 ' m-xY 

a , J 



2* 



a 



17. 



19. 



* 2 + .v a 



14. 
16. 

18. 


y -f- 22° 
1 1 

X X* 

m n 
x* " y* 
1 1 


jc s - f 



x — y x -f- y 

27. Differentials of Irrational Expressions. Let it be re- 
quired to find the differential of the function 

m 
U — X n , 

m and n being positive integers. Raising both members of 
the equation to the nth power, we have 

n n — x m . 

Taking the differentials of both members, 

nu n ~ x du = mx m ~ 1 dx, 



, (") 



whence 






du mx m ~ l m 


x m-l m x m-l m m 


- 1 


dx n u n ~ x n 


I m\n-l — n mn-m — n % 






\X n J X n 





a formula which corresponds to the corollary of Theorem V., 
where the exponent is entire. 

Next, let the fractional exponent be negative. Then 







u = 


m 
X~~" - 


1 

~ ~~ m > 
X n 


and, 


by Th. 


VI., 








du = 


d(x») _ 

2m ' 
X n 


m 

mx* 

n 

X 


dx 

2m 
n 


and, 


for the derivative, 








du 

dx ~ 


x n 

n 







m ---1-, 

= x n dx, 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 39 

From this equation and from (a) we conclude: 
Theorem VII. The formula 

d(x n ) = nx n ~ 1 dx 

holds true whether the exponent n is entire or fractional, posi- 
tive or negative. 

We thus derive the following rule for forming the differen- 
tials of irrational expressions: 

Express the indicated roots by fractional exponents, positive 
or negative, and then form the differential by the preceding 
methods. 

Examples. 

i. d Va -f- x = d(a -f x)* = \(a -f- x) ~ hlx = —. — - — r-. 

4j ( a — (— Xj» 

2. d^-rr = d [b(a + x) - *] = bd(a + *)■-'* 

(a-\-x)* l \ i / ■ j \ • / 

3. d(a + frr')* = |(« + bx*) - i 2bxdx = —JL—dz. 

v ' v 7 (« + &&)* 



EXERCISES. 

Form the differentials of the following expressions: 



I. 


Va + a. 


2. 
5- 
8. 

1 1. 

14. 


V£ - ». 


3- 

6. 

9- 
12. 
15- 


Va — bx. 


4- 


Va - x\ 
a 


Va - bx\ 
b 


Vx + ?/. 


10. 
13- 


Vx + y° 
(a + fc)i 
a; Va + #• 


Va + ££ 3 * 
(a: - «)f. 

x Va — x. 


Va - kc 2 * 
*/ 3 ^« - %'• 



Find the values of — - in the following cases: 
ax 

16. u — mx -| — r 17. u = (#ia? a — ??-)*. 



40 THE DIFFERENTIAL CALCULUS. 

1 8. u == </«^ 4- fo; 3 . IQ. U = 7— „. 

-j- car 

20. ic — x Va — x. 21. u = x Vx 1 -\- a. 

a -\- x a — x 

22. U = . 2%, U =■ ; . 

a — x J a -f x 

28. Logarithmic Functions. It is required to differentiate 
the function 

u = log x. 
We have 

Au = log (:s -f At) — log 2; = log — log (l -j ). 

X \ x / 

It is shown in Algebra that we have 

log (1 + h) = M(h - W + ih 3 - etc.), 
M being the modulus of the system of logarithms employed. 

Hence, puting — for li, we find 

x 







Au 

~Ax~ 


x \ 


1 Ax 

'2 ~x~ 


+ 


1 Ax' 
3 sc" 


- etc. 


and, 


passing to the limit, 














du = 


Mdx 

x ' 






du 

dx 


a;' 



In the Naperian system M = 1. In algebraic analysis, 
logarithms are always understood to be Naperian logarithms 
unless some other system is indicated. Hence we write 

d'log x 1 , n dx 

Example. 

, , d(axy) axdy A- aydx dy , dx 

d-log axy = -- — — = — ! — = = — A . 

J axy axy y x 

Eemaek. We may often change the form of logarithmic 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 41 

functions, so as to obtain their differentials in various ways. 
Thus, in the last example, we have 

log (axy) = log a + log x + log y, 

from which we obtain the same differential found above. The 
student should find the following differentials in two ways 
when practicable. 



EXERCISES. 

Differentiate: 

dx 
i. log (a 4- x). Ans. — ; — . 2. log (x — a). 

v ' a-\- x ° v ' 

3. log (z a + £'). 4. log (x 3 - b). 

5. log mx. 6. log rax 2 . 

7. log («£ n -f- #)• 8. log m 37 . 

9. log (x + #). 10. log (x - y). 

11. log xy. 12. log (£ 2 -f 2/ 2 )» 

13. log (« + 6) y . 14. log^-. 

, a; + « , , « — x 

17. # log a. 18. log (« — x) m . 

29. Exponential Functions. It is required to differentiate 
the function 

u = a x , 

a being a constant. 

Taking the logarithms of both members, 

log u = x log a. 

Differentiating, we have, by the last article, 

dn 
4 ' log u = — = dx log «. 



42 THE DIFFERENTIAL CALCULUS. 

Hence du — u log a dx = a x log a dx' 9 

s = «* log a, 

which is the required derivative. 

If a is the Naperian base, whose value is 

e = 2.71828. 

we have log a = 1. Hence 

d-e x _ x 

dx 

Hence the derivative of e x possesses the remarkable prop- 
erty of being identical with the function itself. 

EXERCISES. 

Differentiate: 
i. a %x . Aiis. 2a' 2x log a dx. 2. a nx . 

7,. C aJrX . 4. (i a + nx e fomx+nv 

6, h mx - v . 7. h~ nx . 8. a x a v . 

9. a x b v . 10. a? x b* v . 11. ah x b~ 2 v. 

12. e x+< \ 13. eV y , 14, e aa: + 6j/ , 

30. The Trigonometric Functions. 

The Sine. Putting h for the increment of x, we have, by 
Trigonometry, 

sin (x -f- h) — sin a; = 2 cos (# -f 2^) sin |&, 

Now, let A approach zero as its limit. Then, 
sin (x -f- //) — sin £ becomes rf sin x; 
h becomes dx, because it is the increment of x; 
cos (x -f- \li) approaches the limit cos x; 
sin \li approaches the limit \h or \dx, because when 
an angle approaches zero as its limit, its ratio to its sine 
approaches unity as its limit (Trigonometry). 
Hence, passing to the limit, 

^•sin x — cos xdx. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS, 43 



The Cosine. By Trigonometry, 

cos (x -f h) — cos x = — sin (x -f- %h) sin ^A. 
Hence, as in the case of the sine, 
d cos x = — sin x dx. 
Taking the derivatives, we have 
d sin x 



dx 

(i'COS x 

dx 



cos x: 



= — sin .7-. 




M N 



Fig. 8. 



Geometrical Illustration. In 
the figure, let OX be the unit- o 
radius. Then, measuring lengths 
in terms of this radius, we shall have 

NK — sin x; MB = sin (x -f- 7*) ; . • . PB = A sin x. 
ON = cos x; OM = cos (x -j- 7i); . * . KP = J cos x. 

Ab'>, supposing a straight line from A" to B, 
PK = - KP = KB sin PBK; 

pb = jcb cos ppa: 

When i? approaches K as its limit, the angle PBK ap- 
proaches XOK, or #, as its limit, and the line KB becomes 
dx. Hence, approaching the limit, we find the same equa- 
tions as before for d sin x and d cos x. 

It is evident that so long as the sine is positive, cos x di- 
minishes as x increases, whence d'eos x must have the nega- 
tive sign. 

The Tangent. Expressing the tangent in terms of the sine 

and cosine, we have 

sin x 
tan x = „ 

COS X 

Differentiating this fractional expression, 

, , cos o^-sin x — sin .t^*cos x sin 2 xdx 4- cos 3 xdx 
d tan x = -^ = -z — 



COS X 



COS X 



== sec 3 xdx, 
which is the required differential, 



44 THE DIFFERENTIAL CALCULUS. 



We find, by a similar process, 

7 , cos x . , dx 

d cot x = cr — — = — esc xdx = r-=— ; 

sm x sin x 

1 d • cos x sin xdx 

a • sec x = cr = = — = 5 — 

COS X cos X cos X 

= tan a; sec xdx; 

d'cosec x = — cot x esc atfo. 



EXERCISES. 

Differentiate: 

1. cos (a -f- y). 2. sin (5 — y). 3. tan (c + #)• 

4. sin y cos 2;. 5. tan u cos v. 6. sin w tan v. 

7. sin «#. 8,.cos«y. 9. tan mz. 

10. sin (A -f- my). 11. cos (^ + ^y)- I2 » sm (^ — m y)< 

13. cos 2 2; • [d'eos 2 x = 2 cos a;^* cos a; = — sin 2xdx], 

14. sin 2 a;. 15. sin 3 y. 16. sin 2 nz. 

sin a: „ sin 2 x cos 2 a; 

17. . 18. . 19. 



cos y cos y sin y 

20. Show that f?(sin a y -f- cos 2 y) = 0, and show why this 
result ought to come out by § 24. 

21. Differentiate the two members of the identities 

cos (a + y) = cos « cos y — sin « sin y, 
sin (a + z) — cos a sin z + sin « cos z, 

and show that the differentials of the two members of each 
equation are identical. 

22. Show that ^*log sin x = cot x dx; 

d'log cos x = — tan a: <^x. 

31o Circular Functions. A circular function is the in- 
verse of a trigonometric function, the independent variable 
being the sine, cosine, or other trigonometric function, and 
the function the angle. The notation is as follows: 

If y = sin z, we write z — sin (_ 1} y or arc-sin y; 
If n = tan x, we write x = tan (_ 1} u or arc-tan u; 
etc. etc. etc. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 45 

Differentiation of Circular Functions. If we have to dif- 
ferentiate 

z = sin (_ 1} y, 
we shall have 



dz=--^=. (b) 



y = sin z; cly = cos z clz ~ Vl — sin 2 z dz; 

rl — sin" 2 y 1 — y 

The Inverse Cosine. If 2 be the inverse cosine of y, we 
find, in the same way, 

cly 

y 

TAe Inverse Tangent. If we have 
z = 'tan (_1) y; 
then, y = ^ an z j ( ty — se ° 2 z dz = (1 -\- tan 2 2)^2; 

7 dy 

1 + r 

The Inverse Cotangent. We find, in a similar way, 

rf .oot<-i j,= __|L.. (d ) 

2^e Inverse Secant. If we have 

2 = sec (_1) ^/; 



then, y = sec z; dy = tan z sec 2 <rfc = y Vy* — 1 dz; 

...* = —&=. w 

y *V - 1 

TOe Inverse Cosecant. We find, in a similar way, 

dy 



d'CSC (-1) ?/ 



y +V - 1 



46 THE DIFFERENTIAL CALCULUS. 





EXERCISES. 


Dif 


ferentiate with respect 


to 


x or 2: 


i. 


sin (_1) ax. 




2. cos (_1) (x-\- a). 


3- 


sin (-1) (mx + a). 




4. cos ( 1} -. 
a; 


5- 


tan<-» (,-!). 




6. tan*-*(f + i). 


7- 


tan <-.,(£ + «). 




8. tan^ 1 )^ 8 ). 


9- 


sec^- 1 ) (2 + ^). 




10. sec (-1) Iz ). 


11. 


sin (_1) fla; cos (_1) — . 




12. sec (-1) x 2 tan (_1) x, 



a 

Kote. — The student will sometimes find it convenient to invert the 
function before differentiation, as we have done in deducing the differen- 
tial of sin C— i) x. 

13. We have, by comparing the above differentials, 

t?(sin _1 y -\- cos -1 y) = 0; 
f?(tan~ 1 y -f- cot ~ * y) = 0; 
^(sec -1 y + csc~ * y) = 0. 

Show how these results follow immediately from the defini- 
tion of complementary functions in trigonometry, combined 
with the theorem of § 24 that the differential of a constant 
quantity is zero. 

32. Logarithmic Differentiation. In the case of products 
and exponential functions, it will often be found that the dif- 
ferential is most easily derived by differentiating the logarithm 
of the function. The process is then called logarithmic dif- 
ferentiation. 

Example 1. Find -~ when y = x" 1 *. 
dx * 

We have log y = mx log x; 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 47 

-d. — m log x dx -j- mdx; 

dv 

— •- = y(m log a; -j- m) — mx mx (l -\- log x). 

ClX 

^ _ sin m x 

Example 2. ?/ = — — . 

J COS n X 

We have log y = m log sin a; — n log cos #; 

dy _ m cos a; » sin a^ 

2/^/a; _ sin x cos a; ' 

^/ sin" 1 "" 1 a:, „ , . . x 

-7- = ■ -—r- (m cos a: -4- n sin £)• 

da; cos n + x x x ' 7 

MISCELLANEOUS EXERCISES IN DIFFERENTIATION. 

Find the derivatives of the following functions with re- 
spect to x: 

1. y = x log x. Ans. -y- = 1 -f- log x. 



2. y = log tan x. Ans. -j- = 



da; sm 2a; 

3. « = log cot x. Ans. -— = : — —. 

J to da; sin 2a 

x . dy « a 

4- y = -777^ ZiV- ^' -±: = 



\/(a* - x") dx (a* - xy 

x n . dy nx n ~ 1 

5- V = Ti-T-rs- ^ w *- 



(1 + a;) n ' *" da; (1 + a;) 7t + 1 * 

e* — e ~ x . dy 4 

*" + «-*" dx (0-4- *-•)■• 



7. y = log (e* + e- a ). -4ws. ^ = ^ _, . 



8. y = log tan — 4- — , . ^4ws. 

* to \4 ^ 2 / da cos x 

x . dy e x (l — x) — 1 

9 - y = *=V Ans - ± "= -(?— ly- 
io . y = ?(i+*)+ rtLz*) ins * = 1 . 

9 tf(l+z)- V(l-«) dx x 4/(1 -a') 



48 THE DIFFERENTIAL CALCULUS. 
TT nl ( f*_ ) " / dy ny 

i 

12. y = tan a S. ^ 5 . ^- = - ^^ log «■•-. 

13. y = 3". ^rcs. — - = ar*(l + log a?). 

• /in a dy cos (log #) 

14. y = sm (log a;). ^iws. -^- = — - — -. 

CI X X 

, _ x a dy 1 

15. y = tan - x — - . ^4«s. -— = — , 

* Vl - a; 2 ^ Vl - a; 2 

i6 - y = to «(r^3*-| t «- 1 ^ 



Ans. ~y~ = ~ 

ax 1 — 



x< 



, /Vl + a; 2 + a; 

I 7 . ?/ = log A / — == . 

V V 1 + x — x 

a dy 1 

Ans ' ^ = -Jt=> 



dx \/\ 



.r 



_ 1 — tan x a dy . . . * 

18. v = . Ans. -rf- = — (cos x 4- sin a;). 

^ sec a; aa; v 

^.y=iog(io ga; ). A ™-% = ^^- 

. I — a; 3 . <fy — 2 

20. w = sin - — ; — 5. Ans. -f- = 3 — . — 5. 

* 1 -f a; 2 da; 1 -f x 

, , /« cos x — ~b sin a; 

2\. y — log y — 7 — . 

3 ° r a cos a; + # sm x 

Ans. cl V- = =^ 

da; « 2 cos 2 a; — # 2 sin 2 a: 

22. If y = -, prove the relation ' - -j , = 0. 

3 Vl + 2/ 4 Vl + a: 4 

23. 2/ =e-a 2 * 2 . ^4^5. -j- = — 2a*xy. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 49 
1 1 



24. y 



(a + x) m {b + xy 



Ans. 



dy_ 
dx 



m(h -f- x) -f- ?z(# -|- #) 
(a + x) m + 1 (b-\-x) n + v 



25. y = (a 2 -f- a; 2 ) tan ~ * — . ^4ws. -p- = 2x tan ~ * — \~ a. 

CI CtX (I 

dy = 1 __ 

da; (1 — sc) 4/1 _ ^ 

2 



!6 - y = /(r^)- ^ n5 - 

17. y = as + logcoB^-aj^««. ^| = j 



28. y = x sin -1 a\ 

29. y = tan a; tan~ * a. 



+ tan a;* 

a dy ■ 1 ■ , 
-4 ?i5. -— = sm -1 a; + - 



4 dy 2xi, tan a; 

ylw5. -=^- = sec a; tan ~ x x + ^ — ; — -,- 

da; ' 1 -f- a- 2 



30. y = sin wa;(sin x) n . 



Ans. 



dy 
dx 



n (sin o;) m_1 sin (n -f- l)a\ 



3i- 2/ = 



(sin nx) m 
(cos wa;) n ' 

.4ws. 



da; 
32. y = e cos rx 



Ans. 



dx 



33. y = log 



34. y = «». 



a -\~l tan - 



cZy _ mn (sin war) m_1 cos {mx — nx) 
(cos wx) n + * 

e - a 2 ** ^a 2 a; cos ra; + r sin ra;). 

dy 



— r*Ans 



ab 



a — I tan 



i«.<t. 



ctx „ „ a; 7 „ . ,, x 
a cos"-- J sin\y. 

dy x x (l — log x) 



dx 



x 4- 1 
35 . y==s i n ->___. 



-4?is. -^- = 



36. y = tan -1 (m tan a*). ^4ws. 



da* 4/1 

<*y _ 



t/a; cos 2 a + w " si 11 * *" 



50 THE DIFFERENTIAL CALCULUS. 

37. y = sec x — - -5 -jr. -4/w. -£- 



38. 



^(a 1 - x 2 ) dx ^(a 2 - x 2 )' 

y = (x + a)tsm- 1 U/~ - ^(ax)J. 



Ans. ^- = tan ^i/iL 
ax * n 



2 


1 + z 2 * 


- tf(a* - b 2 ) 


a + ^ cos x 


2nx n ~ 1 


x 2n -f 1" 


2 


4/(1 - s 3 )* 


1 



dy 

39. # = sin - * 4/ (sin #). ^4w,9. -~ = ^ 4/(1 + cosec a:). 

40. y^tan-'j—^,. ^«-£ = 

, b 4~ a cos x . dy 

41. ?/ = sin -1 — — . Ans. ~- = 

* a -f b cos x dx 

,^-1 . dy 

V'*^"*- 1 *^ AnS 'dx~ = 

_i 1 a dy 

44 . 3, =t an-J^>-i.^. & = __^_ 

33. Derivatives with Respect to the Time. — Velocities. If 
we have a quantity which varies with the time, so as to have a 
definite value at each moment, but to change its value con- 
tinuously from one moment to another, that quantity is, by 
definition, a function of the time. We now have the defini- 
tion: 

If we have a quantity <p, expressed as a function of the 

time = t, the derivative, -jr, is the velocity of increase, 

or rate of variation of <p at any moment. 

This is properly a definition of the word velocity; but it 
may be assumed that the student has already so clear a con- 
ception of what a velocity is, that he needs only to study the 
identity of this conception with that of a derivative relatively 
to t, which he can do by the illustration of § 19. 

The student is recommended to draw a diagram to rep- 
resent the problem whenever he can do so. 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 51 

EXERCISES. 

i. It is found that if a body fall in a vacuum under the in- 
fluence of a constant force of gravity, the distances through 
which it falls in the first, second, third, fourth, etc., second 
of time are proportional to the numbers of the arithmetical 

progression 

1, 3, 5, 7, etc., 

or, putting a for the fall during the first second, the total fall 

will be 

a -\- 3a -\- 5a -{■ Ha -\- etc., 

continued to as many terms as there are seconds. It is now 
required to find, by summing t terms of this progression, how 
far the body will fall in t seconds, and then to express its 
velocity in terms of t, and thus show that the velocity is 
proportional to the time. 

Ans. (in part). The total distance fallen in t seconds will be at*. 
The velocity at the end of t seconds will be 2at. 

2. The above motion being called uniformly accelerated, 
prove this theorem: If a body fall from a state of rest with 
a uniformly accelerated velocity during any time r, and if the 
acceleration then ceases, and the body continue with the uni- 
form velocity then acquired, it will, during the next interval 
r, fall through double the distance it did during the first 
interval. 

Find (1) how far the body falls in r seconds; (2) its velocity at the end 
of that time; (3) how far, with that velocity, it would fall in another 
interval of r seconds; then show that (3) = 2 X (1). 

3. The radius of a circle increases uniformly at the rate of 
m feet per second. At what rate per second will the area be 
increasing when the radius is equal to r feet ? 

Find (1) the expression for the value of the radius r at the end of t 
seconds, and (2) the area of the circle at that time. Differentiate this 

area, and then substitute for t its value in terms of r. Note that ( t = — ) . 

We shall thus have %nmr for the velocity of increase of area. 



52 THE DIFFERENTIAL CALCULUS. 

4. A body moves along the straight line whose equation is 

x — ty = 

with a uniform velocity of n feet per second. At what rate 
do its abscissa and ordinate respectively increase ? 

%n , n 

Ans. ■—=. and — — • 
VS Vb 

5. A man starts from a point h feet south of his door,, and 
walks east at the rate of c feet per second. At what rate is 
he receding from his door at the end of t seconds? 

Ans. If we put u = his distance from his door, we shall 

have 

du _ c*t 
dt~ u' 

6. A stone is dropped from a point h feet distant in a hori- 
zontal line from the top of a flag-staff 9a feet high. At 
what rate is it receding from the top of the flag-staff (1) after 
it has dropped t seconds, and (2) when it reaches the ground, 
assuming the same law of falling as in Ex. 1 ? 

At the end of t seconds the square of the distance from the top of the 
flagstaff — u? = 6 2 + aH*. On reaching the ground we should have 

du_ 54a 2 
dt ~ V6 2 +81a 2 " 

7. The sides of a rectangle grow uniformly, both starting 
from zero, and the one being continually double the other. 
Assuming one to grow at the rate of m feet and the other 2m 
feet per second, how fast will the area be growing at the end 
of 1, 2, 10 and t seconds? How fast, when one side is 4 and 
the other 8 feet ? 

8. The sides of an equilateral triangle increase at the rate 
of 2 feet per second. At what rate is the area increasing 
when each side is 8 feet long ? 

Note that the area of the triangle whose sides = s is ^ / . 



DIFFERENTIATION OF EXPLICIT FUNCTIONS. 53 

9. A man walks round a lamp, 20 feet from it, keeping 
the distance with a uniform motion, making one circuit per 
minute. Find an expression for the rate at which his shadow 
travels on a wall distant 40 feet from the lamp. 

10. The hypothenuse of a right triangle is of the constant 
length of 10 feet, but slides along the sides at pleasure. If, 
starting from a moment when the hypothenuse is lying on 
the base, the end at the right angle is gradually raised up at 
the uniform rate of 1 foot per second, find an expression for 
the rate at which the other end is sliding along the base at 
the end of t seconds, and explain the imaginary result when 
t> 10. 

11. Two men start from the same point, the one going 
north at the rate of 3 miles an hour, the other north-east 5 
miles an hour. Find the rate at which they recede from each 
other. 

12. A body slides down a plane inclined at an angle of 30° 
to the horizon, at such a rate that it has slid 3f feet at the 
end of t seconds. At what rates is it approaching the ground 
(1) at the end of t seconds, and (2) after haying slid 75 feet ? 

13. A line revolves around the point (a, i) in the plane of 
a system of rectangular co-ordinate axes, making one revolu- 
tion per second. Express the velocity with which its intersec- 
tion with each axis moves along that axis, in terms of a, the 
varying angle which the line makes with the axis of X. 

. dx %bn dy 2ci7r 

Am. —r. - 



dt sin 2 a' dt cos 2 a 

14. A ship sailing east 6 miles an hour sights another ship 
7 miles ahead sailing south 8 miles an hour. Find the rate 
at which the ships will be approaching or receding from each 
other at the end of 20, 30, 60 and 90 minutes, and at the 
end of t hours. 



54 THE DIFFERENTIAL CALCULUS 



CHAPTER V. 

FUNCTIONS OF SEVERAL VARIABLES AND 
IMPLICIT FUNCTIONS. 

34. Def. A partial differential of a function of sev- 
eral variables is a differential formed by supposing one of the 
variables to change while all the others remain constant. 

The total differential of a function is its differential 
when all the variables which enter into it are supposed to 
change. 

A partial derivative of a function with respect to a 
quantity is its derivative formed by supposing that quantity 
to change while all the others remain constant. 

Remaek. The adjective partial may be omitted when the 
several variables are entirely independent. 

Example. Let us have the function 

u = x\y -\-z)A- yz. (a) 

Differentiating it with respect to x as if y and z were con- 
stant, the result will be 

da — 2x(y -\- z)dx, (b) 

which is the partial differential with respect to x. Also, 

(£)-«*+•> 

is the partial derivative with respect to x. 

In the same way, supposing y alone to vary, we shall have 

du = (x* + z)dy, (c) 

(*)"* + * 



PARTIAL DERIVATIVES. 55 

which are the partial differential and derivative with respect 
to y. For the partial differential and derivative with respect 
to z we have 

du = (x* + y)dz\ (d) 

'diC 



dzi- x+ y- 

Notation of Partial Derivatives. 1. A partial derivative 
is sometimes enclosed in parentheses, as we have done above, 
to distinguish it from a total derivative (to be hereafter de- 
fined). But in most cases no such distinctive notation is 
necessary. 

2. In forming partial derivatives the student is recom- 
mended to use the form 

D x u instead of -=— j 

because of its simplicity. It is called the D x of n. The equa- 
tions following (b), (c) and (d) would then be written: 

D x u = 2x(y + z); 
D y ii = x* -f z; 
D g u = x* + y. 

EXERCISES. 

Find the derivatives of the following functions with respect 
to x, y and z: 

i. v — x* — xy + y 2 . 

Ans. D x v = 2x — y; D y v = — x + 2y; D z v — Q. 

2. W = X 3 -j- %*y + xz ' 3« U — X*lfz*. 

4. u — x log y + ?/ log a;. 5. w = (cc -j- y -f- z) 7 . 
6. m = |/(ic + my). 7. w = (x -\- %y -f- 32;)*. 

Note. In forms like the last three, begin by taking the 
total differential, thus: 

du = i(as + 2y + 3z)~*d - (x + 2?/ + 3s) 

= i(a + %!/ + 3*)~* (<fo + 2<y + 3<fc)« 



56 THE DIFFERENTIAL CALCULUS. 

Then, supposing x alone to vary, D x u = 

supposing y alone to vary, D y u = 

supposing z alone to vary, D z u = 



2(x+2y+3zf 

1 

(x + 2y+ 3zf 

_3 

2(x+2y+Szf 



8. w={x-\- y + z) n . g. iv = (x* + y 1 + z 2 ) n . 

io. to = cos (mx -\- y). n. w = sin (a; -f- % + 3s). 

i2. v = tan (a: — y). 13. # = sec {mx -j- ws). 

14. # r= cos 2 («.?; + fo). 15. v = e a: + J/ . 

16. u = ze 2 ' + ye 35 . 17. w = £ y -j- y*. 

18. ?i = sin (#+#) cos (re— y). 19. w = x sin y — y sin x. 

35. Fundamental Theokem. T lie total differential of 
a function of several variables, all of lulwse derivatives are 
continuous, is equal to the sum of its partial differentials. 

As an example of the meaning of this theorem, take the 
example of the preceding article, where we have found three 
separate differentials of u, namely, (b), (c) and (d). The 
theorem asserts that when x, y and z all three vary, the re- 
sulting differential of u will be the sum of these partial differ- 
entials, namely, 

du = 2x(y -f z)dx -j- (z 2 -\- z)dy ~\- (x 2 -f y)dz. 

To show the truth of the theorem, let us first consider any 
function of two variables, x and y, 

U = 0(x, y). (1) 

Let us now assign to x an increment Ax, while y remains 
unchanged, and let us call u' the new value of u, and A x n the, 
resulting increment of u. We shall then have 

«/ = <f>{ x + Ax, y); (2) 

A x u =s ${% -f- Ax, y) - <p(x, y). 



TOTAL DIFFERENTIALS. 57 

In the same way, if x retains its value while y receives the 
increment Ay, and if we call A y ii the corresponding incre- 
ment of u, we have 

A v u = <t>{x, y + Ay)- <P(x, y). (3) 

When Ax and Ay become infinitesimal, these increments 
(2) and (3) become the partial differentials with respect to x 
and y. 

Now, to get the total increment of u, we must suppose both 
x and y to receive their increments. That is, instead of giv- 
ing y in (1) its increment Ay, we must assign this increment 
in (2). Then for the increment of u we shall have, instead 
of (3), the result 

A y u' = </>{z + Ax,y + Ay) - <p(x + Ax, y). (4) 

Note that (3) and (4) differ only in this: that (3) gives the 
value of A y u before x has received its increment, while (4) 
gives A y u after x has received its increment, and is therefore 
the rigorous expression for the increment of u due to Ay. 
Now, what the theorem asserts is that, when the increments 
become infinitesimal, the ratio of A y u' to A y u approaches 
unity as its limit, so that we may use (3) instead of (4). To 
show this, let us put 



^y)-=m 



dy 

Then, supposing Ay to become infinitesimal, and putting d y u 
for that part of the differential of u arising from dy, we shall 
have, from (3) and (4), 

d y u = (p'(x, y)dy; (3') 

dy = 4>'{x + Ax, y)dy. (4') 

When Ax approaches zero as its limit, <p'(x + Ax, ?/) must 
approach the limit (p'(x, y), unless there is a discontinuity in 



58 THE DIFFERENTIAL CALCULUS. 

the function <fi f , which case is excluded by hypothesis. Thus, 
using (3') for (4'), we have 

Total differential of u ~du = \--\dx + <p'{x f y)dy 

The same reasoning may be extended to the successive cases 
of 3, 4, . . . 11 variables. 

The following are examples of finding some differential al- 
ready considered in Chap. IV., by this more general process. 

1. To differentiate u = xy. 

du _ du _ 

dx ~ 3> dy ~~ 

Total differential, du — ydx + xdy, 

2. u — — = aw " \ 

V J 

du , t?w- 7 

du — y~ hlx — xy~ 2 dy = 5 — - * 

3. u = ax -j- Zwy -f- ca^z. 

dx = a + y + Cy * ; 

^ , . 

— - z= bx + ca;z; 

5 = «»; 

e??* = (# -|" £?/ -f- cyz)dx -f- (5a; + cxz)dy + cxydz, 
as in § 25, Example 1. 



DIFFERENTIATION OF IMPLICIT FUNCTIONS. 59 



EXERCISES. 

Write the total differentials of the functions given in the 
exercises of § 34. 

36. Principles Involved in Partial Differentiation. All 
the processes of the present chapter are aimed at the following 
object: Any derivative expression, such as 

du n 

Tx> orD * u > 

presupposes (1) that we have the quantity u given, really or 
ideally, as an explicit function of x, and perhaps of other 
quantities; (2) that we are to get the result of differentiating 
this function according to the rules of Chap. IV. , supposing 
all the quantities except x to be constant. 

Now, because it is often difficult or impossible to find u as 
an explicit function of x, we want rules for finding the values 
of D x u, which we could get if Ave had u given as such a func- 
tion of x. For example, we might be able to find the equa- 
tion u = cp(x) if we could only solve one or more algebraic 
equations. If, for any reason, we will not or cannot solve 
these equations, we may still find D x u whenever the equations 
would suffice to give u as a function of x if we only did 
solve them. The following articles show how this is done in 
all usual cases. 

37. Differentiation of Implicit Functions. Let the rela- 
tion between y and x be given by an equation of the form 

<P(x, y) = 0. («) 

Representing this function of x and y by 0, simply, and 
supposing for the moment that x and y are independent 
variables, so that need not be zero, we shall have, by the 
last section, 

deb — -^dx + -j-dy. 
^ dx dy * 



60 THE DIFFERENTIAL CALCULUS. 

But, introducing the condition that equation (a) must be 
satisfied, dtp must be zero, because x and y must so vary as to 
keep tp constantly zero. We then find, from the last equation, 

^0 

dy _ dx D x (f> .. 

dx~ dcp~ Rrf) 9 * ' 



which is the required form in the case of an implicit function 
of one variable. 

Cor. If from an equation of the form x = f(y) we want to 
derive the value of D x y, we have 

</>{x, y) =x-f(y) = 0; 
dtp _ dtp _ df(y) _ dx 
dx ~ ' dy dy dy' 



Hence 



dx dx' 
dy 

Example. To find D x y from the equation 

<t>{*j y) = y — ax = 0. 



We have 



dtp 



dtp _, dy 



dx ' dy dx 

the same result which we should get by differentiating the 
equivalent equation y = ax. 

Remark. If we should reduce the middle member of (1) by clearing 
of fractions, the result would be the negative of the correct one. This 
illustrates the fact that there is no relation of equality between the two 
differentials of each of the quantities x, y and <p, all that we are concerned 
with being the limiting ratios dy : dx; dtp : dx, and d<p : dy, which limit- 
ing ratios are functions of x and y. 

We may, indeed, if we choose, suppose the two dx's equal and the two 
dy's equal. But in this case the two dtp's must have opposite algebraic 
signs, because their sum, or the total differential of tp, is necessarily zero. 
Now, if we change the sign of cither of the dtp's, we shall get a correct 
result by a fractional reduction. 



DIFFERENTIATION OF IMPLICIT FUNCTIONS. 61 



EXERCISES. 

Find the values of -j-, -7- or -=— from the following equa- 

UX U/Z aX 

tions: 

1. y 2 — ax — 0. 2. y 2 — yx -}- x 2 = 0. 

3. z a + te: + z 2 = 0. 4. u(a-x)+t( 2 (b + x)=0. 

5. log 2 + log y = c. 6. log (x+y) + log (z-#) = c. 

7. sin a; -J- sin y = c. 8. sin arc — sin #7/ = c. 

9. m + ^ sin u = a;. 10. x (1 — e cos 2) = #. 

38. Implicit Functions of Several Variables. The pre- 
ceding process may be extended to the case of an implicit 
function of any number of variables in a way which the 
following example will make clear. 

Let u be expressed as a function of x, y and z by the 
equation 

a* + xu 2 + (x 2 + y 2 )u + x z + y 3 + z 3 = 0. 

Since this expression is constantly zero, its total differential 
is zero. Forming this total differential, we have 

(3u 2 + 2xu + x 2 + y f )dto + (u 2 + 2wz + 3z a )<fo 

+ (2wy + 3?/ 2 )d?/ + Sz 2 dz = 0. 

By § 34 we obtain the derivative of u with respect to x by 
supposing all the other variables constant; that is, by putting 
dy = 0, dz = 0, and so with y and z. Hence 

du _. w 2 + 2wa? + 3# 2 

= Z4w = — 



dx x Su 2 + 2ux + x 2 + if 

du n _ 2wy -f 3?/ 2 

^- - ^V* - ~ 3,,- + %ux + a; 9 + y 8 

du -. Sz 2 

Dm 



dz * 3u 2 + 2ux-^-x 2 + y 2 



62 



THE DIFFERENTIAL CALCULUS. 



EXERCISES. 

Find the derivatives of u, v or r with respect to x, y and z 
from the following equations: 

i. xn 3 -\- y 2 u 2 -\- z*u = x 2 yz. 

2. a cos (x — u) -\-b sm (x -\-u) = y. 

3 . n x + u v = u\ 4. r w + v + r x ~ v = r z . 

5. v log x -\- z log v = y. 6. e r cos a; + e* cos y = e v . 

7. u 2 — 2tix cos 2 -f # 2 = a 2 . 8. t> 2 -j- 2ttz cos z ^- x 2 — V. 

39. Ctose of Implicit Functions expressed by Simvlta- 
neoux Equations. If we have two equations between more 
than two variables, such as 

F x (u, v, x, y, etc.) = 0, F 9 (u, v, x, y, etc.) = 0, 

then, if values of all but two of these variables are given, we 
may, by algebraic methods, determine the values of the two 
which remain. We may therefore regard these two as func- 
tions of the others, the partial derivatives of which admit of 
being found. 

In general, suppose that we have n independent variables, 
x lf x 2 . . . x n , and m other quantities, u l9 u 2 . . . u mi connected 
with the former by m equations of the form 



\(u l9 u % . 


. . u m} x 1} x 2 . 


.x n ) = 


\(u l9 u, . 


. . u m , x xi x 2 . 


. . = 



F m {u l9 u. 



%i> x i 



x n ) = 0. 



(«) 



By solving these m equations (were we able to do so) we 



should obtain the m u's in terms of the 



n xs 



in the form 



U x — (pX X ,y X i • - ■ X n)\ 



tl m = <Pm( x i? x *' •• X n)? 



<*) 



DIFFERENTIATION OF IMPLICIT: FUNCTIONS. 63 
and by differentiating these equations (b) we should find the 

mn values of the derivatives - T - 1 : -~: . . . -^- 2 : etc. 

dx x dx 2 ax x 

Now, the problem is to find these same derivatives from (a) 
without solving (a). 

The method of doing this is to form the complete differen- 
tial of each of the given equations (a), and then to solve the 
equations thus obtained with respect to du l9 du a , etc. 

The results of the differentiation may, by transposition, be 
written in the form 



dF x . ,dF 1/7 


. dF x _ 
'+d^ m du ~ = 


dF x , 


— etc. ; 


dF> , . dF m 7 . 
-7— - du. + -r- 1 du 9 4- . . 
du x l du 2 8 


. dF * 7 
'+du m du ~ = 


Fd 2 . 
~ ~dx~ x *** 


— etc.; 



dF m , dF m dF m -. dF m 

dUi + du , + .. . + dUm= - dXi _ etc. 



By solving these m equations for the m unknown quantities 
du x , du 2 . . . du m) we shall have results of the form 

du x = M x dx x -f M % dx % + . . . + M n dx n ; 
du q = N x dx x + iV/7^ 2 + . . . + N n dx n \ 
etc. etc. etc. etc.; 

where M x> N l9 etc., represent the functions of u x . . . w^, 
2 X . . . # n , which are formed in solving the equations. 
We then have for the partial derivatives 

du, lir du, ,^ , 
-j- 1 = M.\ -i~ = Mj etc. 
dx, l dx n 



Example. From the equations 

r cos 6 — #, 

r sin 

it is required to find the derivatives of r and with respect 
to x and y. 



e = x '\ («') 



64 THE DIFFERENTIAL CALCULUS. 

By differentiation we obtain 

cos ddr — r sin Odd = dx; 
sin ddr -j- r cos ddd = dy. 

Multiplying the first equation by cos d and the second by 
sin 6, and adding, we eliminate dd. Multiplying the first by 
— sin 6 and the second by cos 6, and adding, we eliminate dr. 
The resulting equations are 

dr = cos 6dx -f- sin ddy; 
rdd = cos 6dy — sin ddx. 

Hence, as in the last section, 



\dx) 


C0S 6 > (| 


)- 


sin 0; 


fdff\_ 
\dxl 


sin (dd 

r ' \d~y 

EXERCISES. 


)= 


cos 

7* 


. From the equations 








r sin d = x — 


*/> 





r cos 6 = x -\- y, 

find the derivatives of r and with respect to x and ?/. 
2. From the equations 

^e u = r cos #, 
ue~ v — r sin 6>, 

find the derivatives of u and v with respect to r and 6. 



Am. (Jr)=-H^in0 +C -«cos0); 



FUNCTIONS OF FUNCTIONS. 65 

3. From the equations 

if -f- ru = x % -f- ?/% 



rw = zy 



find the derivatives of r and w with respect to x and y. 
4. From the equations 5. From 

u 2 — 2wz cos 6 -f- z 1 — # 2 , 
u? -f 2uz cos 6 -j- z 2 = Z> 2 , 



2 3 


+ ^ + * 3 


— 2#?/z = 


= o, 






% + y + z = 


= 0, 


find 


dz 
clx 


and 


dy' 





fi , du du dw chv 
lz> d¥ ; ~dz~' d0 m 

40. Functions of Functions. Let us have an equation of 
the form 

u =f(<t>, f, 6, etc.); (a) 

where 0, ip, 6, etc., are all functions of x, admitting of being 
expressed in the form 

=/,(*); *=/.(*); »=/,W; etc. (8) 

If any definite value be assigned to x, the values of 0, ip, 
6, etc., will be determined by (b). By substituting these val- 
ues in {a), u will also be determined. Hence the equations 
(a) and (I?) determine u as a function of x. 

By substituting in (a) for 0, if:, 6, etc., their algebraic 
expressions f (x), f(x), etc., we shall have u as an explicit 
function of x, and can hence find its derivative with respect 
to x. But what we want to do is to find an expression for 
this derivative without making this substitution. 

By differentiating (a) we have 

, du j , . du 7I . du jn , 
du = -r-idch 4- -rrdib 4- -777"^ 4- etc. 
d0 dty> r «# 

By differentiating (b), 

deb = V-flfo: rtM = -Afo: d# == -j-dx; etc. 
dx dx dx 



66 THE DIFFERENTIAL CALCULUS. 

By substituting these values in the last equation and divid- 
ing by dx, we have 

du __ du d(p du d>/j du dd 

dx d(p dx d>p dx dd dx ' . 

The significance of this equation is this: a change in x 
changes u in as many ways as there are functions <fi, tp, 6, etc. 

■77 -rr-dx is the change in u through d>: 
dtp dx ° ° 

— -j-dx is the change in u through tp; 

etc. etc. 

The total differential is the sum of all these separate 
infinitesimal changes, and the derivative is the quotient of 
this total differential by dx. 

EXERCISES. 

i. Find -7— from the equations 

ctx 

u = a sin (mv -f- w) -\- b sin (mv — w)\ 

v = c -\- nx'y w = c — nx. 

We find — - = am cos (mv + w) + &m cos (my — w>); -— = n; 
dv ' eta 

— - = a cos (m» + «o) — o cos (ma — w); — - — — n\ 

dw . dx 

Whence, by the general formula, 

-— = an(m — 1) cos (mv + w) + bn(m -f- 1) cos (ttiv — w). 

2. Find -7— from 

u = e* + <^; 

= ^; ip = ne~ x . Ans. e*+ x - ne*~ x . 

-i. Find -7- from 

v 2 + vcp -|- ?// = a; 



<f> = m(a + y); ip = wy. 



FUNCTIONS OF FUNCTIONS. 67 



4. Find -r=- from 
dz 



r cos x — r sm x = a — y; 

z = mz -f ^; y = cos ^2. 

«;. Find -p from 
D dz 

r 3 + xt* + #V + 3 = 0; 

x 2 -\- az = 0; y 3 -\- az 2 = 0; = nz. 

41. The foregoing theory applies equally to the case in 
which the function is one of two or more variables, some of 
which are functions of the others. For example, if 

u = (p{x, z), (a) 

then, whatever be the relation between x and z, we shall 
always have, for the complete differential of u, 

Suppose that x is itself a function of z. We then have 

dx = -x- dz. 
dz 

By substitution in the first equation we have 



du 



'(dn\dx , (du\~~\ -. 



du __ (du \ dx (du \ ( . 

m ''dz~ \dx) "dz + \dz)' W 

du 
The two values of -=- which enter into this equation are 

ClZ 

different quantities. A change in z produces a change in u 
in two ways: first, directly, through the change in z as it 
appears in ((/); second, indirectly, by changing the values of 

x in (a). The first change depends upon [-=-] in the second 



68 THE DIFFERENTIAL CALCULUS. 

(du \ dx 
-t- ') -j-; while the first mem- 
ber of (I)) expresses the total change. 

It is in distinguishing the two values of a derivative thus 
obtained that the terms partial derivative and total derivative 
become necessary. If we have a function of the form 

u=f( x , y> w • • • *)> 

in which any or all of the quantities x 3 y, w, etc., may be 
functions of z, then the partial derivative of u with respect 
to z means the derivative when we take no account of the 
variations of x, y, w, etc.; and the total derivative, with 
respect to z, is the derivative when all these variations are 
taken into account. 

In such cases the partial derivative has to be distinguished 
by being enclosed in parentheses (§ 34). This is why the last 
equation is written 

dx 



du __ fdu\ , (du\ dx 
dz ~ \dz) \dx I dz 



42. Extension of the Principle. The principle involved 
in the preceding discussion may be extended to the case of 
any number of independent variables and any number of 
functions. If we have 

r = <f)(u, v, id . . . x, y, z . . .), 

in which x, y, z, etc., are the independent variables, while 
u, v, w, etc., are functions of these variables, we shall have 

Then, since u, v, iv, etc., are functions of z, y, %, etc., we 

have 

du 7 , du 7 . , 

dv = -j-dx + --dy + etc. 
dx dy J . 



FUNCTIONS OF FUNCTIONS. 69 

By substituting these values in the preceding equation we 
find* 

*- [©+© 'I +©£+•• ■> 
+ [(f) + (2) | + ©*+•••> 

+ 

Hence, writing r for 0, its equivalent, 

tfr _ /dfr\ f dr\du fdr\dv 
dx ~ Vlx) + fej^" + [dvldx + 6tC,; 
etc. etc. etc. etc. 

EXERCISES. 

The independent variables r and 6 being connected with x 
and y by the equations 

x = r cos d, 
y = r sin 6, 

it is required to find the derivatives of the following functions 
of x, y, r and 6 with respect to r and 6. We call each of the 
functions u. 



i. u — r 2 -{- 2xy cos 26. 






Here we have 






© = * 




(f) = - 4 ^ sin 


^- = 3 ,cos 29; 




— = 2.# cos 20; 
dy 


j— = cos 0; 




d5r 


T/T = — f Sill = - 
<20 


-v, 


—- = r cos = 



* Here, when we use the symbol </> instead of r, there is really no 
need of enclosing the partial derivatives in parentheses. We have done 
it only for the convenience of the student. 



70 THE DIFFERENTIAL CALCULUS. 

du _ ldu\ du dx du dy 
dr ~ \dri ' dx dr dy dr 

= 2r + 2y cos cos 20 -f- 2x sin cos 20 
= 2?-(l -f cos 29 sin 20) = r(2 + sin 40); 

and, in the same way, 

— = 2r 2 cos 40. 
dO 

We might have got the same result, and that more simply, by sub- 
stituting for x and y in the given equation their values in terms of r and 
0. But in the case of implicit functions this substitution cannot be 
made; it is therefore necessary to be familiar with the above method. 

a 2 , x 2 — y 2 na 
2 . u = — , + = cos 20. 
r a 

« 2 , 6 2 2«6 
3- « = ? + p--p-. 

4. w = r 2 — (a; — ?/) 2 . 

1 

5 * u ~ x sin 20 + y cos 20* 

1 1 

6. w 



a cos 2(9 ?/ sin 20' 

7. m = r 3 + ^ 3 — y 3 - 

Let v and w be given as implicit functions of p and 6 by 
the equations 

w = av; 



(a) 

v * -j- w 2 = 2/) sin 0. J w 

It is required to find the total derivatives of the following 
functions with respect to p and respectively: 

8. u = v 2 -\- w 2 — p 2 . 9. u — v 2 — 2vw cos -j- w 2 . 

ao f . v . n 

10. 11 = - — . 11. u = (v -\- w) sm 0. 

WW 

12. U = (V — w) COS 0. 

13. «, = w 2 — v 2 -f 2(zo + v)p cos 0. 



PARTIAL DERIVATIVES. 71 

From the pair of equations (a) we find 



dp 2p' dp 2p' 

— yr = \v cot 0; — n - z\w cot 6; 
ao as) 

which values are to be substituted in the symbolic partial derivatives of u. 

43. Remarks on the Nomenclature of Partial Derivatives. 
There is much diversity among mathematicians in the no- 
menclature pertaining to this subject. Thus, the term "par- 
tial derivative" is sometimes extended to all cases of a deriva- 
tive of a function of several variables, with respect to any one 
of those variables, though there is then nothing to distinguish 
it from a total derivative. 

Again, Jacobi and other German writers put the total deri- 
vatives in parentheses and omit the latter from the partial 
ones, thus reversing the above notation. 

If we have to express the derivative of cp(x, y, z, etc.) with 
respect to z, the English writers commonly use the symbol 

-=- in order to avoid writing a cumbrous fraction. We thus 
az 

have such forms as 

dl* 1 . xy . y* 



dx\a 



(a? xy tf\ 

W "*" V ^ c*J 



D[ x l + W + t\ 



each of which means the derivative of the expression in paren- 
theses with respect to x, and which the student can use at 
pleasure. 

44. Dependence of the Derivative upon the Form of the 
Function. Let x and y be two variables entirely independent 
of each other, and 

u = (p(x, y) (a) 

a function of these variables. Without making any change 
in u or x, let us introduce, instead of y, another independent 



72 



THE DIFFERENTIAL CALCULUS. 



variable, z, supposed to be a function of x and y. Then, after 
making the substitution, we shall have a result of the form 

u = F{x, z). (b) 

Now, it is to be noted that although both u and x have the 

same meaning in (b) as in («), the value of — will be differ- 

O/X 

ent in the two cases. The reason is that in ((f) y is supposed 
constant when we differentiate with respect to x, while in (b) 
it is z which is supposed constant. 
Analytic Illustration, Let us have 

u = ax* -j- by*. 
du 



This gives 



dx 



= 2ax. 



(c) 



Let us now substitute for y another quantity, z, determined 
by the equation 

z = y -\- x or y — z — x. 

We then have u = ax* -j- b(z — x) 2 ; A 

— - — 2ax -{- 2b(x — z); 

which is different from (c). 

Our general conclusion is: The partial derivative of one 
variable with respect to another depends not only upon the re- 
lation of those two variables, but upon their relations to the 
variables which we sup- 
pose constant in differen- 
tiating. 

Geometrical Illustra- 
tion. Let r and 6 be the 
polar co-ordinates of a 
point P, and x and y its 
rectangular co-ordinates. 
Then fio. 9. 

x = r cos d; 
y = r sin 6; 
r* = x* + y\ 




(d) 



PARTIAL DERIVATIVES. 73 

Regarding r as a function of x and y, we have 

dr x a . 

-7- = - = cos 6. (e) 

ax r v 

But we may equally express r as a function of x and #, thus: 

r = x sec 0. (/) 

We then have -j- = sec ^. (g) 

Referring to the figure, it will be seen that we derive (e) 
from (d) by supposing x to vary while y remains constant; 
that is, by giving the point P an infinitesimal motion along 
the line PQ || to OX. In this case it is plain that the incre- 
ment of r (SQ) is less than that of x. But in deriving (g) 
from (/) we suppose x to vary while 6 remains constant. 
This carries the point P along the straight line OPE; and 
now it is evident that the resulting increment of r (PR) is 
greater than that of x. 



74 



THE DIFFERENTIAL CALCULUS. 



CHAPTER VI. 
DERIVATIVES OF HIGHER ORDERS, 
45. If we have given a function of x, 

y = 00*), 



we may, by differentiation, find a value of 



dy 
dx' 



This value 



will, in general, be another function of x, which we. may call 
<p'(x). Thus we shall have 



dx 



4>'(x). 



Now, this function 0' may itself be differentiated. If we 
call its derived function <p n ', 
we shall have 

dy 



d 



dx __ dcp'{x) _ „ 



dx 



dx 



= <P"{x).(a) 



M 



Let us examine the geo- 
metrical meaning of this 
equation, by plotting the 
curve representing the origi- 
nal equation y = (x). 

Let x f x' and x" be three 
equidistant values of the ab-"~ 
scissa, so that the increments 
x' — x and x" — x' = Ax are 
equal. Let P, Q and R be 
the corresponding points of the curve. Let y, y f and y" 
be the three corresponding values of y. 




AX 



Xo 



X2 



Fig. 10. 



DERIVATIVES OF HIGHER ORDERS. 75 

Then we may put 

Ay = y' — y = MQ, 
. A'y^y" -y'^NR, 

as the two corresponding increments of y. 

It is evident that these increments will not, in general, 
be equal; in fact, that they can be equal only when the three 
points of the curve are in the same straight line. If D is the 
point in which the line PQ meets the ordinate of E, then 
DR will be the difference between the two values of Ay, so 
that we shall have 

DR — A'y — Ay = increment of Ay. 

Hence, again using the sign A to mark an increment, we 
shall have 

DR = A Ay = A'y, (I) 

in which the exponent does not indicate a square, but merely 
the repetition of the symbol A. 

Theorem I. Wlien Ax becomes infi?iitesimal, A'y becomes 
an infinitesimal of the second order. 

For, if D be the point in which PQ produced cuts the 
ordinate XJi, we shall have, in the triangle QRD, 

DB=QD <™jm =A * y , (5) 

v sin QRD u v ' 

When Ax becomes an infinitesimal of the first order, so do 
both QD and the angle RQD, but the angle QRD will remain 
finite, because it will approach the angle QDN as its limit. 
Hence the expression will contain as a factor the product of 
two infinitesimals of the first order, and so will be an infini- 
tesimal of the second order. 

Since both the quantities QD and RQD depend upon Ax, 
we conclude that the ratio 

Ay 

Ax' 
may remain finite when Ax becomes infinitesimal. In fact. 



76 THE DIFFEBENTIAL CALCULUS. 

from the way we have formed these quantities, we have 

A*v Ax dx 

lim. -— = lim. — ^- = — r - = d>"(x). 
Ax* Ax dx ^ v ' 

Hence — 

Theokem II. If we take tiuo equal consecutive infinitesimal 
increments, = dx, of the independent variable, then — 

1. The difference between the corresponding infinitesimal 
increments of the function divided by dx* will approach a 
certain limit. 

2. This limit is the derivative of the derivative of the 
function. 

Defi The derivative of the derivative is called the second 
derivative. 

The derivative of the second derivative is called the third 
derivative, and so on indefinitely. 

Notation. The successive derivatives of y with respect to 
x are written 

&y_. <£y. <£y. p+p . 

dx 9 dx*' dx" etC *' 
or D x y; DJy; D x 3 y; etc. 

46. Derivatives of any Order. The results we have 
reached in the last article may be expressed thus: If we have 
an equation 

y = <f>(z), 

the first derivative is given by the equation 

Then, by differentiating this equation, we have, by the last 
theorem, 



DERIVATIVES OF HIGHER ORDERS. 77 

Again, taking the derivative, we have 

and we may continue the process indefinitely. 

EXERCISES AND EXAMPLES. 

i. To find the successive derivatives of ax 3 . 

d / = Sax'; 

ax 

d'y a 

W = 6ax; 

d 3 y a 

and all the higher derivatives will vanish. 

Form the derivatives to the third, fourth or ^th order of — 

2. ax\ 3. bx" 1 . 4. (a -\- x) 3 . 

5. (a — x)\ 6. (a + x)- 3 . 7. (a — x)~ s . 

8. (cf + xy. 9. 2aV + a\ 

10. a -\- bx -\- ex 2 -f- hx 3 -f- lex*. 

11. 1 -f x + x* + x 3 + ^ + a 5 + . . . + x n . 

12. 1 — a; + x 2 — x 3 -f .t 4 — a: 5 -J- . . . -f (- l) n a:'\ 

1 3 n 

13. a' 2 ". 14. xK 15. (r? -f .r) n . 16. (a-|-a) 2 . 

17. If y = e z , find D z n y = a e (\og a) n . 

18. From y — me z , find the nth. derivative. 

19. From y = me hz show that D~ n y = h n y. 

Find the first three derivatives of the expressions: 

20. z z . 21. ax x . 22. x ax . 

23. log a. 24. log (a -\- x). 25. m log .t. 

26. log (0 — #). 27. log (a -j- w»). 2tS - log (^ — m-0- 

29. Show that if y = sin x, then — ^ = — y. 

(Vy _ _ g " <- * y _ dry 
~dtf ~ y ' dx n + n ~ dx n ' 



78 THE DIFFERENTIAL CALCULUS. 

30. Show that the same equations hold true if y — cos x 
or if y — a cos x -j- b sin x. 

31. Find the law of formation of the successive derivatives 
of sin mx and cos mx. 

Especially, the (n -f- 4)th derivative = nth der. x what? 
(n -f 2)th derivative = nth der. X what? 

32. Find the nth derivative of e mx . 

3$. Find three derivatives of e mz sin nz. 

34. If u = y y , show that ^ = (1 -f log y)^ + ^. 

35. Find two derivatives of w = tan z. 

36. Find two derivatives of u = cos 2 z. 

37. Find two derivatives of u — sec 2 z. 

3S. Find two derivatives of u — cos 2 2 — sin 2 z. 

39. Find two derivatives of u = cos 2z. 

40. Find two derivatives of u = e~ x *. 

41. Find two derivatives of u = sin (_1) a;. 

47. Special Forms of Derivatives of Circular and Ex- 
ponential Functions. Because 

cos x = sin {x -\- \rt) and — sin x = cos (x -{- -k^), 

the derivatives of sin x and cos x may be written in the form 

D x sin x = sin (x -f ^-7r) 
and D x cos a; = cos (x -\- ^n). 

Hence, the sine and cosine are such functions that their 
derivatives are formed by increasing their argument by \n. 
Differentiating by this rule n times in succession, we have 

d n sin x 



DJ 1 sin x = 



dx r 



^ m d n cos x 

D x n cos x — — =-- — = cos 
dx n 



= sin (x + ^7r); 



results which can be reduced to the forms found in Exercises 
29 and 30 preceding. 



DERIVATIVES OF HIGHER ORDERS. 79 

48. Sttccessive Derivatives of an Implicit Function. If 
the relation between y, the function, and x, the independent 
variable, is given in the implicit form 

f{x, y) = 0, 

then, putting u for this expression, we have found the first 
derivative to be 

du 
dy _ dx . 

dx ~ du* ^ ' 

dy 

The values of both the numerator and denominator of the 
second member of this equation will be functions of x and y, 
which we may call X t and Y s . We therefore write 

&=-£i. (b) 

dx y; K) 

Differentiating this with respect to x, we shall have 

_ y -— 4- X — -' 
d*y ' dx ' dx . . 

X 4 and Y t being functions of both x and y, we have (§ 41) 
dX, = ldX\ + (dXJ\dy 



dx \ dx J \ dy )dx y 
dx \ dx I \ dy I dx 



dii 
Substituting in these equations the values of -— from (b), 

and then substituting the results in (c), we shall have the re- 
quired second derivative. 

The process may then be repeated indefinitely, and thus 
the derivatives of any orders be found. 

Example. Find the successive derivatives of y with re- 
spect to x from the equation 

x 1 — xy -j- if = u — 0. 



80 THE DIFFERENTIAL CALCULUS. 

Wehave J = Zx - y; ^t=- x + 2 r , 

dy _ 2x — y m 
dx ~ x — 2y' 

which is a special case of (a) and (b), and where 

X t — 2x — y and Y t = — x -f- 2y. 

Differentiating the equation (a'), we have 



(«') 



(*- 


-z») d{%x - y) - ^ 


d(x - 2y) 
y) dx 


(X- 


(x - 2yy 
" 3 4-|) + ^- 


■4I- 1 ) 






(x - 2yY 

dii 
Substituting the value of — from (a'), we have 

cix 

(Ty = (s-2y)(-3;/)+3:r(2s-y) 

^ 2 (a — 2#) 3 

_ 6 (a; 2 — 27/ -f «/ 2 ) _ 6w 

(^-%) 3 " = (x - 2yf 

EXERCISES. 

• Find by the above method the first two or three derivatives 
of v with respect to x, y or z, from the following equations: 

/ x a d*v 2(a 4- v) 

i. zv — a(v — z). Ans. -^- — ~ r{. 

v ' dz {a — z) 

2. v*y -J- vy 3 — ^' 

3. y 2 -f ^ -f if = I. 

4. v(a — x) 2 -\- v 2 (b -f- x) = c. 

5. log (v + z)+ log (V - 2?) = €. 

6. sin w« — sin ny = 7j. 

7. i'(l — a cos z) = A. 

8. If u — e sin w = g, show that 

<^ 2 w _ 1 — e 
dedg — (1 — e cos u)*' 



DERIVATIVES OF HIGHER ORDERS. 81 

49. Leibnitz's Theorem. To find the successive deriva- 
tives of a product in terms of the successive derivatives of its 
factors. 

Let uv = p be the product of two functions of x. By suc- 
cessive differentiation we find 

dp _ dv du m 

dx ~ dx dx' 

d 2 p _ d 2 v du dv cVu 

dx* dx 2 dx dx dx 2 ' 

cVp _ d 3 v du d 2 v Jl*v> dv d 3 u 

dx 2 ~ dx 3 dx dx 2 dx 2 dx dx 3 

So far, the coefficients in the second member are those in 
the development of the powers of a binomial. To prove that 
this is true for the successive derivatives of every order, we 
note that each coefficient in any one equation is the sum of 
the corresponding coefficient plus the one to the left of it in 
the equation preceding. Now, let us have for any value of n 

d n p d n v , du d n ~H , , , x 

-tit — ii-t-- 4- n-. — ^— - — < -f- etc. ; (a) 

dx 11 dx n ] dx dx 11 - 1 ' ' v ' 

the successive coefficients being 

1; n; [£); (^); etc. (Comp. § 6.) 

Then, in the derivative of next higher order the coefficients 
will be 

i; » + i; £) + - or (-+I); 



and, in general, 



or 



m 



d n + 1 v 
That is, ^ — -£ is formed from (a) bv writing n + 1 for n. 

dx n + 1 v ' " 

Hence, if the rule is true for n, it is also true for n -j- 1 . But 
it is true for n — 3; . * . for n — 4, etc., indefinitely. 



82 TEE DIFFERENTIAL CALCULUS. 

50. Successive Derivatives with respect to Several Equi- 
crescent Variables. Studying the process of § 45, it will be 
seen that we supposed the successive increments of the inde- 
pendent variable to be equal to each other, and to remain 
equal as they became infinitesimal, while the increments of 
the functions were taken as variable. This supposition has 
been carried all through the subsequent articles. 

Def. A variable whose successive increments are supposed 
equal is called an equicrescent variable. 

We are now to consider the case of a function of several 
equicrescent variables. 

If we have a function of two variables, 

u = cp(x, y), 

the derivative of this function with respect to x will, in 
general, be a function of x and y. Let us write 

^=cp x (x,y). 

Now, we may differentiate this equation with respect to y 
with a result of the form 

,du 

d dx~ 

-^=<p x , v {x,y). 

Using a notation similar to that already adopted, we rep- 
resent the first member of this equation in the form 

d 2 u 
dxdy' 

In the D-notation this is written 

D\ v u. 

In either notation it is called ''the second derivative of 
u with respect to x and y." 

As an example: If we differentiate the function 

u = y 2 sin (mx — ny) (a) 



DERIVATIVES OF HIGHER ORDERS. 83 

with respect to x, and then differentiate the result with 
respect to y, we have 

l) x u = — = my cos (mx — ny); 

G/X 



D* XtV u = 1 ^ 1 = 2my cos (mx — ny) -j- mmf sin (mx — uy). 



d*n 
dxdy 

51. We now have the following fundamental theorem: 
d 2 a d*u m 
dxdy dydx 7 
or, in words, 

The second .derivative of a function with respect to two 
equicrescent variables is the same whether toe differentiate in 
one order or the other. 

Let u = <p(x, y) be the given function. Assigning to x 
the increment Ax, we have 

Au _ <fi(x 4- Ax, y) — <p(x, y) 



Ax Ax 



a) 



Au 

In this equation assign to y the increment Ay, and call A— 

the corresponding increment of -p. Then the equation will 

AiX 

give 

^at + a — = ^ x + Ax > y + J ^ ~ ^ Xy y + J // ) 

Ax Ax Ax 

Subtracting (1) and dividing the difference by Ay, we 
have 

dx _ <p{x -\- Ax, y + Ay) - <p(x, y -f Ay) - <p(x + Ax, y) + <p(x, y) 
Ay AxAy 

The second member of this equation is symmetrical with re- 
spect to x and y, and so remains unchanged when we inter- 
change these symbols. Hence we have 

A Au A Au 
Ax Ay 



Ay ' Ax 



84 THE DIFFERENTIAL CALCULUS, 

for all values of Ax and Ay, and therefore for infinitesimal 
values of those increments. Thus 

jlu _ du 

Cl-Y- d'-r- 

dx dy 



dy dx 

or B\, y u = D 2 VtX u, 

as was to be proved. 

As an example, let us find the second derivative of (a) in 
the reverse order. We have 

— — = 2y sin (?nx — ny) — ny 1 cos {mx — ,ny)\ 

d 2 n 

— — - = 2my cos (ww — ny) + wm?/ 2 sin (nix — ny); 

(lilu/X 

the same value as before. 

Cokollary. The result of taking any number of succes- 
sive derivatives of a function of any number of variables is 
independent of the order in ivhicli ice perform the differentia- 
tions. 

For, by repeated interchanges of two successive differentia- 
tions, we can change the whole set of differentiations from 
one order to any other order. 

If we have I differentiations with respect to x, m with re- 
spect to y, n with respect to z, etc., and use the D-notation, 
we express the result in the form 

D x l D y m D z n . . . 0. 

Here the symbol D y m means D y D y , etc., m times. 

In the usual notation the same operation is expressed in 

the form 

d i + m + n + ... ( p 

dx l dy m dz n ..." 

The corollary asserts that, using the D-notation, we may 
permute at pleasure the symbols DJ, D y m , D z n , etc., without 
changing the result of the differentiations. 



DERIVATIVES OF HIGHER ORDERS. 85 

EXERCISES. 

Verify the theorem D x D y u = D y D x u in the following cases: 

r. u = x sin y -j- y sin x. 2. u — a; 77 . 

3. u = x log y. 

4. w = « sin (:c -\- y) — i sin (x — y). 

Differentiate each of the following functions once with re- 
spect to z, twice with respect to y, and three times with re- 
spect to x, in two different orders, and compare the results. 

5. ^Hjp. 6. ax\fz\ 

7. a; sin y -\- y sin 2; -f~ z sm ^- 8. sin (Ix -f- wy + W2; ). 

1 



9. If w = 



i\> 



V(x* + 2/ 2 + *') 



I 



, ,, . d*u t d*u . d*u 

show that _+_ 5 + _ J = . 

52. Notation for Powers of a Differential or Derivative. 
Such an expression as du? may be ambiguous unless defined. 
It may mean either 

Differential of square of u; i.e., cl(u 9 ); 

or Square of differential of u; i.e., (du)*. 

To avoid ambiguity, the expression as it stands is always 
supposed to have the latter meaning. To express the differ- 
ential of the square of u we may write either 

d'u* or d(u'), 

of which the first form is the easier to use. 

The square of the derivative -r- may be written either 



fduV r (M 
\dx I dx 7 ' 



86 THE DIFFERENTIAL CALCULUS. 



CHAPTER VII. 

SPECIAL CASES OF SUCCESSIVE DERIVATIVES. 

53. Successive Derivatives of a Power of a Derivativec 
Let us have to differentiate the derivative 



(duV 
\dx) 



with respect to x. 

In such operations the D-notation will be found most con- 
venient. 

Applying the rule for differentiating a square, the result is 



,fdu\ 
\dx) 



-\ , d a 



jdu dx _ 9 du d*u 



dx dx dx dx dx 7f 

or, in the D-notation, 

D x (D x uY = 2D x uDJu. 
In the same way, we find 

d'(D x tc) n (du\ n - 1 d i u , n ln in9 

-V 1 = n Kdx) m = < B ** B * u > 

d'(D x u) n (duV- 1 d'u ,_ . .„ 

d'Wuy = 2 dju^u = 

dx dx" dx s 



SPECIAL CASES OF SUCCESSIVE DERIVATIVES 87 



EXERCISES. 

Write the derivatives with respect to x of the following ex- 
pressions, y being independent of z when it is written as an 
equicrescent variable: 



7- -7--. • «. T7-, • 9- 



• (!)'• 


2 ' \dy) ' 


• »(J)' 




• WW ' 


» (£)•• 


du du 


/oV\ a ow 


dz dy' 


Wa?/ aty" 


(dyVfduV 
' \dz) \d~zJ ' 


• (duV/d'uy 
I4 ' \dx) \dtf) ' 


' \dx) [dx 9 ) ' 


I7 - w) ' 


\dx'J \dx z ) 


du dv du dv 
dx dy dy dx' 



fduV 
; * \dyl ' 
. fduV 

f d'u y 

\dxdyl ' 
(duVfduV 
' \dx~J \dy) ' 
d^uVldS 
dy 2 ) \dy* 



16. ~ -^-J . 17. hr-r . 18 



10. — -— . 11. 1-— J — -. 12. 

Of/ 

(dyy 

I9, lovj lo#J ' 2 °* dxly~ dylz 2I ' IdyV * 

54. Derivatives of Functions of Functions. Let us have, 
as in § 40, 

u=f-(f), (1) 

where ^ is a given function of #. It is required to find the 
successive derivatives of u with respect to z. We may evi- 
dently reach this result by substituting in (1) for ip its ex- 
pression in terms of x, and then differentiating the result by 
methods already found. 

But what we now wish to do is to find expressions for the 
successive derivatives without making this substitution. To 
do this, assign to x the infinitesimal increment dz. The re- 
sulting infinitesimal increment in ip will be 

dip = -r~dx. 
r dx 



88 THE DIFFERENTIAL CALCULUS. 

This, again, will give u the increment 



du 
di 



du = ^dtp, 



or, by substituting for dtp its value, and passing to the de- 
rivative, 

du _ du dip 

dx dtp dx' 

This is a particular case of the result already obtained in 
§40. The second member of (2) is a product of two factors. 
The first of these factors is formed by differentiating a func- 
tion of tp with respect to t/r, and is therefore another (derived) 
function of ip; while the second is, for the same reason, a 
function of x. 

Differentiating (2) with respect to x by the rule for a prod- 
uct, we have* 

du 
d?u _ dtp dtp du d?ip 
dx" 1 dx dx dtp dx* * * 

du 
Now, because — isa function of tb, its derivative with re- 
dtp 

spect to x is to be obtained in the same way as that of u. 

If we put, for the moment, 

we have, as in (2), 

du' _ du* dtp _ d*u dtp _ 
dx ~ dtp dx dtp* dx 9 

JLu 

% 
* The student should note that the expression — — cannot be put in the 

iPu 
form ; - , because the latter form presupposes that ip and x are two in- 
dipclx 

dependent variables, which is here not the case. In fact, u does not con- 
tain x except in i/f. 



SPECIAL CASES OF SUCCESSIVE DERIVATIVES. 89 

and hence, by substitution in (3), 

d a u _ d*u(dipV , du d*tp m .. 

dx~* ~ dipAdx~) ^ dip dx 7 ' W 

which is the required expression for the second derivative. 

From this we may form the third and higher derivatives 
by again applying the general rule embodied in (2), namely : 

If tp is a function of x, ice find the derivative of any func- 
tion, u, of tp by differentiating u with respect to tp, and mul- 

tiplyi7ig the resulting derivative by -j-. 

From the equation (4) we have 

dhi 
<£u _ fdipV W , 2 <£u dip_ cfrp 
dx 3 ~ \dxl dx dtp 3 dx dx 2 

1 du 
d^ip dip du d*ip 
+ dx~* ~dx + # W 

By the rule just given, we have 

dip* _ d 3 u dtp^ 
dx dip* dx* 

■j m du 

dip _ d*u dip 
dx ~ dip* dx ' 

Hence, by substitution and aggregation of like terms, 



d 3 u _ d 3 u (dipV d^u d 2 ip dtp du d 3 ip . . 

dx* ~ dip 3 \dxl W &* d% # dx T ' W 

Repeating the process, we shall find 

d*u _ d*u(dip\ 4 d 3 u d*tp (dipV 

dx* ~ dip\dx) + dip 3 dx* \dx) 

In r d*tp # f ( ripy-] du ( rip 

^ dip % [_ dx 3 dx "r \dx> I J + dify dx* ' K } 



90 THE DIFFERENTIAL CALCULUS. 

Example. Let us take the case of 

u == sin tp, 

tp being any function whatever of x. We may then form the 
successive derivatives as follows: 

du __ du dip _ dtp m 

dx ~ dtp dx ~ dx ? 

d*u . ,fdtp\\ ,d 3 tp 



dx* ^\dx) ' ~~" r dx* 

d'u JdipX* a . ,dipd'tp 

__ = _ cos ^j_ 2sm?/ ,___ r 

. ,dtpd*tp , ,d'tp 

— sin ^>y- -y-f + cos ^?-=-£- 
r dz dx* T dx 






EXERCISES. 

Putting = a function of £, find the first three derivatives 
of the following functions of with respect to x: 

I. U = COS 0. 2. U = J 

3. W = 0\ 4. U = W . 

5. w = log 0. 6. % = efi . 

7. m = sin 20. 8. w = cos 20. 

55. Change of the Equicrescent Variable. Let the relation 
between y and x be expressed in the form 

* = <%)> (i) 

and let it be required to find the successive derivatives of y 
with respect to x, regarding the latter as the equicrescent. 
We may do this by solving (1) with respect to y, and then 
differentiating with respect to x in the usual way. 

But the method of the last article will enable us to express 
the required successive derivatives of y with respect to x in 
terms of those of x with respect to y, which we can obtain 



SPECIAL CASES OF SUCCESSIVE DERIVATIVES. 91 

from (1). By differentiating (1) as often as we please, we 
have results of the form 

D y x - <p'y = x'; j 
D y *x=c?>"y = x"; [ (2) 

D v s x= <p'"y = x"'. ) 
etc. etc. 

x' ', x" , x nr , etc., thus representing functions of y. 
From § 37, Cor., we have 

^L = — =-. (3) 

dx D v x x r K ' 

To obtain the second derivative, we have to differentiate x' , 
a function of y, with respect to x (§ 54). Thus 

d*y _ 1 dx' dy^ 
dx 2 ~ x' % ' dy dx' 

From ^ % = %=*"■ 

From this equation and (3) we have 

(Tx 

dx' ~ ' x h ~ (dx_Y 
\dy) 
Differentiating again, we find 



(£y _ /3z" <W _ 1 efo"\ dy 
dx 3 \ x fi dy x n dy 1 dx 



x 



Jd'xV dx d*x 
_ %x' n - x'x'" _ \dyV "dydy* 

'dz\ l 



(*?Y 
\dyl 



The above process may be carried on to any extent. But 
many students will appreciate the following more elegant 
method of obtaining the required derivatives. 

Imagine that we have solved the equation (1) so as to 
obtain a result in the form 

y = F{x). (5) 



92 THE DIFFERENTIAL CALCULUS. 

If in this equation we substitute for x its value (1), we shall 
have a result in the form 

y = F{<py), (6) 

which, of course, will really be an identity. 

But we may still differentiate (5) with respect to y, regard- 
ing z as a function of y given by (1), by the method of §§ 40 
and 54. Thus we shall have 

d 3 y _ cPyfdxV d*y d 7 x dx dy d*x 
dy 3 ~ dx 3 \dy J dx 3 dy 2 dy dx dy 3 ' 

etc. etc. etc. etc. 

But from the identity (6) y — y, which is obtained from 
(5), we have 

^--1- ?!-0- --^-O- etc 
dy " ' df ~ ° ? dy 3 ~ °' etC * 

Therefore, substituting for the derivatives of x with respect 
toy the expressions x', x" } etc., in (4), we have the equations 

X dx ~ 1? 
X dx^ X dx~ V > 

dx 3 ' dx 1 dx ' 

x >* ^K + 6s'V ^ + (4zV" + 3z" 2 ) ^ + ^4- = 0. 
dz 4 ' c?a; 3 ' v ' dx 7 dx 

Solving these equations successively, we shall find the values 
of -~, ~, etc., already obtained. 

56. Case of Two Variables Connected ly a Third. The 
case is still to be considered in which the relation between x 



SPECIAL CASES OF SUCCESSIVE DERIVATIVES. 93 

and y is expressed in the form 

y = ^(w); x = a (w). (1) 

From these equations it is required to find the successive 
derivative of y with respect to x. 

The first derivative is given by the equation 

dy_ 
dy __ du_ _ D u y 
dx ~ dx ~ D u x 

du 

From the manner in which the second member of this equa- 
tion is formed, it is an explicit function of u alone. Hence 
(§ 54) we obtain its derivative with respect to x by taking its 

du 

derivative with respect to u, and multiplying by -7—. Thus 

dx d 2 y dy d*x 
dx' 



du du* 


du du* du 


\d 


x\ 2 ' dx 
u ) 


dx d*y 
du du 2 


dy d 2 x 
du du 2 



(dxV 
\du ) 



This, again, being a function of u, further derivatives with 
respect to x may be obtained by a repetition of the process. 

EXERCISES. 

Find the second derivative of x with respect to y, and also 
of y with respect to x, when the relation of x and y is given 
by the following equations: 

1. x = a cos u; y = b sin u, 

2. x = a cos 2u; y = h sin u. 

3. x = a cos 2u; y — £(cos u — sin u). 

4. x = u — e sin u; y = u -f e sin u. 

5. x = e u ; y = ue 2u . 



94 THE DIFFERENTIAL CALCULUS. 

6. Show that if 

,, d*u 2 sin u 

y — e u cos u, then - r ^- = -5-7 : u . 

* dy e 2M (cos w — sin u) 

7. Show that the wth derivative of x n -{- ax n ~ 1 -{- dx n ~ 2 is 
n\, n being a positive integer > 1. 

8. Show that 

D x '(u 3 ) = 3u*D x 5 u + V&uD x uD*u + 6(D x u)\ 

9. Show that if v = w n , then 

A, 3 *; = ?iu tl - 1 D x 3 u + 3rc(w — l)u n ~ 2 D x uD x 'u 

+ rc(rc - 1) (w - 2)u n ~\D x u)\ 

10. If u = a cos wza + # si n m ^j show that 

A> + m'w = 0. 

Then, by successively differentiating this result, show that, 
whatever the integer n 9 

D x u + 2 u + m'D x n u = 0; 
D x n+i u -m*D x n u = 0. 

11. If u = e x cos x and v = e x sin x, then 

D x u = — 2v and D x v = 2u. 

Also, ZX> + 4=v = 0; 

D x *u + 4w = 0. 

12. If u = e nx cos ma; and v = e** sin mx, 

show that the successive derivatives of u and may always be 
reduced to the form 

DJu = -4*w — -#<#; TV** = ^ t v -f B&> ( a ) 

where A and B are functions of m and n. Also, find the 
values of A if A 2 , B 1 and i? 2 , and show by differentiating (a) 
that 

^,+1 = 4^,-5^ B t + 1 = B x A t + A x B+ 



DEVELOPMENTS IN SERIES. 95 



CHAPTER VIM. 
DEVELOPMENTS SN SERIES. 

5*7. A series is a succession of terms all of whose values 
are determined by any one rule. 

A series is called 

Finite when the number of its terms is limited; 

Infinite when the number of its terms has no limit. 

The sum of a finite series is the sum of all its terms. 

The sum of an infinite series is the limit (if any) which the 
sum of its terms approaches as the number of terms added to- 
gether is increased without limit. 

When such a limit exists, the series is called convergent. 

When it does not exist, the series is called divergent. 

To develop a function means to find a series the limit of 
whose sum, if convergent, shall be equal to the function. 

We may designate a series in the most general way, in the 
form 

W i + U <x + U z + • • • + U n + Un + l + • • • > 

the nth. terms being called u n . 

58. Convergence and Divergence of Series. No universal 
criterion has been found for determining whether any given 
series is convergent or divergent. There are, however, a great 
number of criteria applicable to a wide range of cases. Of 
these we mention the simplest. 

I. A series cannot be convergent unless, as n becomes in- 
finite, the nth term approaches zero as its limit. 

For if, in such case, the limit of the terms is a finite 
quantity a, then each new term which we add will always 



96 TEE DIFFERENTIAL CALCULUS. 

change the sum of the series by at least a, and so that sum 
cannot approach a limit. 

As an example, the sum of the series 

1 — 1 + 1 —1 + 1 — 1, etc., ad infinitum, 
will continually change from + 1 to 0, and so can approach 
no limit, and so is divergent, by definition. 

II. A series all of whose terms are positive is divergent 
unless nu n = when n = oo . 

To prove this, we have first to show that the harmonic 
series 

i + i + i + |- + etc, ad infinitum, 

is divergent. To do this we divide the terms of the series, 

after the first, into groups, the first group being the 2 terms 

-J + \, the second group the following 4 terms, the third 

group the 8 terms next following, and, in general, the ?zth 

group the 2 n terms following the last preceding group. We 

shall then have an infinite number of groups, each greater 

than \, 

Now, if, for all the terms of the series after the nXh, we 

have 

nu n > a (a being any finite quantity), 

then u n > — , 

and u a + u m+1 +...> «(1 + — ^ + — Lj +...). 

Because the last factor of the second member of this equa- 
tion increases to infinity, so does its product by a, which 
proves the theorem. 

III. If the terms of a series are alternately positive and nega- 
tive, continually dimmish, and approach zero as a limit, 
then the series is convergent. 

Let the series be 

u x — u a + u 9 — u 4 + u 6 — . .. . . 
Then, by hypothesis, 

u x > u % > u 3 > u t > . . . . 



DEVELOPMENTS IN SERIES. 97 

Let us put 8 n for the sum of the first n terms of the series, 
n beiug any even integer, and S for the limit of the sum, if 
any there be. Then this limit may he expressed in either of 
the forms 

S= S n +(w„ + i - w„ + 2 ) + (w tt + 3 — u n + i ) + . . . 
and 

S= S n + 1 — (u n + 2 — u n + 3 ) — (u n+i — u n+5 ) — . . . . 

Since all the differences in the parentheses are positive, by 
hypothesis it follows that, how many terms soever we take, 
the sum will always be greater than S n and less than S n+1 . 
The difference of these quantities is u „ + 19 which, by hypothe- 
sis, approaches zero as a limit. Since the two quantities S n 
and S n+1 approach indefinitely near each other from opposite 
directions, they must each approach a limit S contained be- 
tween them. 

Graphically the demonstration may be shown to the eye 
thus: Let the line 0# 6 represent the sum S n , when n = 6, 

O Se Ss Sio— S Sn S-9 S"r 



TT 



Fig. 11. 

or any other even number; OS, the sum S n , etc. Then every 
succeeding even sum is greater than that preceding, and 
every succeeding odd sum is less than that preceding, while 
the two approach each other indefinitely. Hence there must 
be some limit S which both approach. 
An example of such a series is 

of which the nth. term is — ^ ~. We shall hereafter see 

2m — 1 

that the limit of the sum of this series is -\rt. If we divide 

the terms into pairs whose sums are negative, the series may 

be written 

2 2 2 

etc. 



3'5 7-9 11-13 



98 THE DIFFERENTIAL CALCULUS. 

Pairing the terms so that the sum of each pair shall be posi- 
tive, the series becomes 

2,2 2 2 

3 + W + 9-11 + Wl5 + etC ' 

"We may show by the preceding demonstration that these 
series approach the same limit. 

IV. If j after a certain finite number of terms, the ratio of 
two consecutive terms of a series is continually less than a cer- 
tain quantity a, tvhich is itself less than unity, then the series 
is convergent. 

Let the ?zth term be that after which the ratio is less than 
a. We then have 

u n+1 < au n ; 

u n + 2 < au n + 1 < a 2 u n ; 



Taking the sum of the members of these inequalities, we 
have 

M» + i + tt„ + 8 + M„ + s+ . . . < (tf + a: a + a 5 ~\- . . . )u n . 
But a -f- a % -j- a* + . . . is an infinite geometrical progres- 
sion whose limit when a < 1 is , a finite quantity. 

Hence, putting S for the limit of the sum of the given 

series, we have 

8 < S " + Y^~a Un ' 

The second member of this inequality being a finite 
quantity which S can never reach, S must have some limit 
less than that quantity. 

As an example, let us take the exponential series 

T 2 X 3 

< = l + , + s + 8l+ .... 



DEVELOPMENTS IN SERIES. 99 

\ X 

The ratio of the (n 4- l)st to the Wth term is -. This 

t n 

ratio becomes less than unity when n > x, and it approaches 

zero as a limit. Hence the series is convergent for all values 

of x. 

Cokollaby. A series 

a Q + a x x -f a 2 x* + a 9 z 3 + . . . 

proceeding according to the powers of a variable, x, is conver- 
gent when x < 1, provided that the coefficients a n do not in- 
crease indefinitely. 

Remarks. — (1) Note that, in applying the preceding rule, it does not 
suffice to show that the ratio of two consecutive terms is itself always 
less than unity. This is the case in the harmonic series, but the series is 
nevertheless divergent. The limit of the ratio must be less than unity. 

(2) If the limit of the ratio in question is greater than unity, the series 
is of course divergent. Hence the only case in which Rule IV. leaves 
a doubt is that in which the ratio, being less than unity, approaches 
unity as a limit. But most of the series met with come into this class. 

(3) The sum of a limited number of terms of a series gives no certain 
indication of its convergence or divergence. If we should compute the 
successive terms in the development of e — 10 ° we should soon find our- 
selves dealing with numbers having thirty digits to the left of the deci- 
mal-point, and still increasing. But we know that if we should continue 
the computation far enough, say to 1000 terms, the positive and negative 
terms would so cancel each other that in writing the algebraic sum we 
should have 42 zeros to the right of the decimal-point. 

On the other hand, if the whole human race, since the beginning of his- 
tory, had occupied itself solely in computing the terms of the harmonic 
series, the sum it would have obtained up to the present time would have 
been less than 44. For 1000 million of people writing 5000 terms a day 
for 2 million of days would have written only 10 19 terms. It is a theorem 
of the harmonic series, which we need not stop to demonstrate, that 

s.=4+|- + ^ + --- + ;r <Naplosre - 

t? 4. tvt i imq comm. log 10 19 19 AO „ Q 

But Nap. log 10" = 4343 ° = m ^ 7 = 43.78, 

and yet the limit of the sum of the series is infinite. 



100 THE DIFFERENTIAL CALCULUS. 

59. Maclaurin's Theorem. This theorem gives a method 
of developing any function of a variable in a series proceed- 
ing according to the ascending powers of that variable. 

If x represents the variable, and the function, the series 
to be investigated may be written in the form 

<P{x) =A + A x x + Atf + A 3 x> + . . . ; (1) 

the series continuing to infinity unless is an entire func- 
tion, in which case the two members are identical. 

Whether the development (1) is or is not possible depends 
upon the form of the function 0. Most functions admit of 
being so developed; but special cases may arise in which the 
development is not possible. Moreover, the development will 
be illusory unless the series (1) is convergent. Commonly this 
series will be convergent for values of x below a certain mag- 
nitude, often unity, and divergent for values above that mag- 
nitude. What we shall now do is to assume the development 
possible, and show how the values of the coefficients A may be 
found. 

Let us form the successive derivatives of the equation (1).. 
We then have 

0(z) = A + A x x + A^ + etc.; 

<f/(x) = A x -f 2A 2 x + 3A 3 x' + . . . ; 



tf0 

dx 
d 2 cp 
dx' 

£& 

dx 3 



4>"(x) = l-2A t + 2SA 2 x + 3-4jy - 
4>'"{x) = V2-3A u + 2-3-44tX+ . . . 



j£= <P (n) (x) =l;2-3-4. . . nA n + etc. 

By hypothesis these equations are true for all values of x 
small enough to render the series convergent. Let us then 
put x = in all of them. We then have 



DEVELOPMENTS IN SERIES. 101 

0(0) = A ; .'. -4. = 0(0). 

0(0) =A X ; .-.± = 4/(0). 

<P"(0) = V2A 2 ; ..„^ 2 = I L0- ( O). 
0"'(O) = l-2-3^; ...^ 3 = r -1^0'-(O). 



0™(O) = nlA n ; . • . 4. = ^0 (n) (O). 

By substituting these values in (1) we shall have the re- 
quired development. Noticing that the symbolic forms 0'(O), 
0"(O), etc., mean the values which the successive derivatives 
take when we put x = after differentiation, we see that the 
coefficients are obtained by the following rule : 

Form the successive derivatives of the given function. 

After the derivatives are formed, suppose the variable to be 
zero in the original function and in each derivative. 

Divide the quantities thus formed, in order, by 1; 1; 1*2; 
1'2*3, etc., the divisor of the nth derivative being n\ 

The quotients will be the coefficients of the powers of the 
variable in the development, commencing with the zero power, 
or absolute term. 

EXAMPLES AND EXERCISES. 

r. To develop (a -f- x) n = u in powers of x. We have 
u = (a + x) n ; . • . A = a n . 

■£ = n(a + x)"- 1 ; .-. A, = na n ~\ 

— r = n(n - 1) (a + z) w " 2 ; ■'■A= ^.g ; « n ~ 8 - 



/7 s ^/ 

"^ = n<<n - x ) • • • (^ ~ 5 + x ) (^ + *) 



102 THE DIFFERENTIAL CALCULUS. 

Thus the development is 
{a + x) n = a n + na n -*x + (^)a n ~ V + (fV n_V + • • • » 

which is the binomial theorem. 

2. Develop (a — x) n in the same way. 

3. Develop log (1 -f- x). 
Here we shall have 

£ = rr-* = < 1 + *>-> 

g = 1-2(1 + ,)-; 

etc. etc. 

Noticing that log 1 = 0, we shall find 

log (1 + x) = X — ix* + iiB 8 — ix* + . . . . " 

4. Develop log (1 — #). 

5. Develop cos x and sin a. 

The successive derivatives of sin x are cos x, — sin x, — cos a?, sin x, 
etc. By putting x = 0, these become 1, 0, — 1, 0, 1, 0, etc. Thus we 
find 

. X 2 , X* X 6 , 

BM. = l- ri + ffl ..-,|j+.... ! 

a? 3 , # 5 a? 7 , 
am * = «_-+- -- + .;.-.. 

6. Develop e% where e is the Naperian base. 

x 1 x 3 
Ans. e x = 1 + x + -j + -, + . . . . 

7. Develop e -a: . 

8. Show that 

9. Deduce e sinx = 1 + iC + 2~If+ 



DEVELOPMENTS IN SERIES. 103 

to. Develop sin (a + x) and cos (a -\-x) and thence, by com- 
paring with the results of Ex. 5, prove the formulae for the 
sine and cosine of the sum of two arcs. Find first 

x* x 3 

sin (a + x) = sin a (1 — ~ + . .) + cos a (x — ^--f . .). 

ii. Develop (1 -f e x ) n and show that the result may be re- 
duced to the form 

i + 2 ^ 2 2 1.2^ 2 3 3! ^ p 
i2. Develop e x sin x and e* cos a; and deduce the results 

— « , r*. X . n % . X r\ X 

/v> 2 /v 4 /v 5 

e X COS ^ = 1 + iC — 2 -r — 4 -rr — 4 -r + . . . 

2! 4! 5! 

13. Develop cos 3 x. 

Begin by expressing cos 3 x in the form \ cos 3x -f- £ cos #. 

14. Develop tan ( ~%. 

This case affords us an example of how the process of de- 
velopment may often be greatly abbreviated. It has been 
shown that 

^•tan ( "% 1 



dx 1 + x* 

Now assume 



1 — x* + x* - x° + etc. (a) 



tan (- % = A -f A x x + Aj? -j- etc. 
This gives 

*' t ^" "* = A +-»^» + *Af + etc. (J) 

Comparing (a) and (5), we have 

^i = i; A = - i; A = i; A = - to etc - 

and ^t, = ^4 4 = A 6 . . . =0. 

The value of A is evidently zero. Hence 

tan ( -% = x — ix s + \x b - ^a; 7 + etc. (c) 



104 TEE DIFFERENTIAL CALCULUS. 

15. Develop sin (_1) z. 

SmCe — di = (1 ~ X) ~ > 

we may develop the derivative and proceed as in the last ex- 
ample. We shall thus find 

. ( _ 1} x . 1 x z IS x b . 1-3-5 x 1 . . 

sm x = T + 2*3- + 1** + 5TS 7 + etc * 

60. Ratio of the Circumference of a Circle to its Diameter. 
The preceding development of tan (-1) z affords a method of 
computing the number n with great ease. The series (c) 
could be used for this purpose, but the convergence would be 
very slow. Series converging more rapidly may be obtained 
by the following device: 

Let a, a', a" ', etc., be several arcs whose sum is 45° = \n. 
We then have 

tan (« + «' + a" + etc.) = 1. 

Let t, f, t" , etc., be the tangents of the arcs a, a' ', a" , etc. 
If there are but two arcs, a and a' , we then have, by the 
addition theorem for tangents, 

i±|_ = l; or t + f = l-tt'. 

If there are three arcs, a, a' , and <*", we replace t' by 
t' + f 



in the last expression, and thus get 
t + t' + t"- tt't" = l-tt'- t't" - tt". 
We now have to find fractional values of t, f and t" of the 

form — , m being an integer, which will satisfy one of these 

equations. Unity is chosen as the numerator because the 
powers of the fraction are then more easily computed. The 
simplest fractions which satisfy the last equation are 

/ — 1. /' — z.. /" — L 



DEVELOPMENTS IN SERIES. 



105 



We then have, from the development of tan (_1) t, etc., 



_1 1 , 1 

a ~2 3-2 3i ~5-2 5 



a 



ft 



5 35 3 ' 5-5 ! 
111 



— n = a A- a f A- a". 
4 

These series were used by Dase in computing 7t to 200 
decimals. 

A combination yet more rapid in ordinary use is found by 
determining a and a' by the conditions 



tan a = 



5' 



4a — a f = — n. 
4 



"We then have 



tan 2 a 
tan 4 a 



2t 
1-f 
120. 
119 ; 



o 
12 ; 



and because a' = ^a — \n = 4a- — 45°, we have 

, _ tan 4a- — 1 _ 1 
tan a ~ tan 4a + 1 ~ 239* 

Hence we may compute it thus : 

1 



1 X . 1 



3-5 



1 



a- = — — — 



5'5 5 7-5' 



+ 



+ 



239 3'239 3 ' 5'239 6 ? 

n = 4a — a'. 

Ten or eleven terms of the first series, with four of the 
second, will give n to 15 places of decimals. 



106 THE DIFFERENTIAL CALCULUS. 

61. In developing functions by Maclaurin's theorem we 
may often be able to express the derivatives of a certain order 
as functions of those of a lower order. The process of find- 
ing the higher derivatives may then be abbreviated by retain- 
ing the derivatives of lower orders in a symbolic form, so far 
as possible. 

EXAMPLES. 

i. Let us develop 

u = log (1 + sin x) = <fi(x). 

We now have 

. . . N cos x 1 — sin x 

(b'(x) = - — ■ — ; = = sec x — tan x; 

v ' 1 -}- sin x cos z 

<p"(x) = sec x tan x — sec 2 x = — sec x<p\x). 

Now, in continuing the differentiation, we use the last of 
these forms instead of the middle one. Thus 

(p'"{x) = — sec x tan x <p'{x) — sec xcf)"{x) 
— — sec x tan x (p'(x) -j- sec 2 xcp'(x) 
= - <P\x)4>"{x). 
We may now find the successive derivatives symbolically. 
Omitting the symbol x after 0, we have 

V = - 0'0 iv - 30" 0'"; 

vi = — 0'0 V — 40"0 iv — 30'" 2 . 



etc. 




etc. 


Supposing x = 0, 






0(0) = 0; 




0iv( O ) = _ 2; 


0'(O> = + 1; 




V (O) = + 5; 


<f>"(0) = - 1; 




0vi (O) = _ 16 . 


0"'(O) = + 1; 




etc. etc. 


ence 

log (1 -|- sin x) = x — 


x 7 
2 


a: 3 z 4 z 5 a: 6 
f 6 12 + 24 45 



DEVELOPMENTS IN SERIES. 107 

2. To develop u = tan x. 

Let us write the equation in the implicit form 

u cos x — sin x = 0. 

Then, by differentiation and division by cos x, we find 

D x ti = 1 -{- u 1 ; 

D*u = 2uD x u = 2u + 2^ 3 ; 

DJu = 2wA> + 2(D x tiY; 

D x u = 2uD x *u + 8D x uD x u + 6(ZVw) a - 
Putting w — 0, we find the even derivatives to vanish and 
the odd ones to become 1, 2, 16, etc. Hence 

tan x = x -f %x 3 + f T x 6 + . . . . 

3. To develop u = sec x. 

Differentiating the form u cos x — 1 = 0, we find 

D x u cos x — u sin x = 0. («) 

The successive derivatives of this equation may each be 
written in the form 

M cos x — N sin x = 0. (#) 

For, if we differentiate this equation with respect to x, it 
becomes 

(D X M - N) cos x - {M + i)^) sin x = 0. 

Hence the derivative of (h) may be formed by putting 

M' = D X M -NT; N' = M + D X N, (c) 

and writing M' and JV"' instead of M and iV in the equation. 
In (a) we have 

M = D x u; JSf=u. 

Then, by successive substitution in (c), 

M' = D x \l- u; JST = 2D x u; 

M" = D x \l- 3D x ti; N" = 3D x *u - u; 

W" = D x *u- 6D x *u + u; N'" = ±DJti - 4:D x u; 
M iy = D x b u-10D/u + 52>««5 -iV^ iv = 5ZV™ - 10D x *u + w. 
Jf v = D x 6 u — l$D x l n + 157), V, - u; 



108 



THE DIFFERENTIAL CALCULUS. 



When x = 0, we have sin x = 0, cos x = 1, u = 1, and 
hence M = M f = ... — in all the equations. Thus we 
find, for x = 0, 

D*u = u = l; 

D x *u = 6-1 = 5; 

D x 6 u = 75 - 15 + 1 = 61; 

etc. etc. ; 

while the odd derivatives all vanish. Hence 



sec x 



61 



1+3^+^+^+.. 



4! 



6! 



62. Taylor's Theorem. Taylor's theorem differs from 
Maclaurin's only in the form of stating the problem and ex- 
pressing the solution. The problem is stated as follows: 

Having assigned to a variable x an increment h, it is re- 
quired to develop any function of x -\- li in powers of h. 

Solution. Let be the function to be developed, and let 

us put 

u = <p(x); ) ' 
u' = <p{x+Ji). ) 



(1) 



Assume 

u> = X + XJi + X,¥ + XJi 3 + etc (2) 

where X , X l9 etc., are functions of x to be determined. 
Then, by successive differentiation, we have 

flv' 

^- = X x + 2XJi + 3XJi* + ±XJi 3 + etc. 



dh* 
cVuf_ 
dh 3 
etc. 



= 2X 2 + 2'3X 3 h + 3-4X 4 /* a + etc.; 

= 1-2-3X, + 2'3'4:X i h + etc. 
etc. etc. 



(3) 



We now modify these equations by the following lemma: 
If we have a function of the sum only of several quantities, 

the derivatives of that function with respect to those quantities 

will be equal to each other. 



DEVELOPMENTS IN SERIES. 109 

For if in f(x -f- h) we assign an increment Ah to x and to 
h separately, the results will be f(x -f- h + ^) and /(z -f- z/A 
-f- A), which are equal. 
It follows that we have 

du' _ du' 
dh ~~ dx' 
Now these equal derivatives, like u' itself, are functions of 
x -\- h alone, so the lemma may be applied to as many suc- 
cessive derivatives as we please, giving 
d\i' __ d 2 u' 
dh' ~ dx' ' 
d 3 u' _ d 3 u' 
dh 3 ~ dx 3 ' 
etc. etc. 

Now let the derivatives with respect to x be substituted for 
those with respect to h in equations (3), and let us suppose h 
to become zero in equations (2) and (3). Then u' and its de- 
rivatives will reduce to u and its derivatives, and we shall get 



-1. 


= u; 
du 
~dx~> 


x, 


1 <Tu m 
~ V2 dx*' 


^3 


1 d 3 u 


~ 1-2-3 dx 3 ' 


x„ 


_ 1 d n u 
~ n\ dx n ~' 



Then, by substitution in (2), we shall have, for the required 
development, 

, , du h , d'u h* , d 3 u h 3 

u = u + -j- =- + yir^+ji ., + etc. 
dx 1 ate 8 1-2 dz 3 1 • 2 • 3 

This formula is called Taylor's Theorem, after Brook 
Taylor, who first discovered it. 



110 THE DIFFERENTIAL CALCULUS. 



EXAMPLES AND EXERCISES. 



i. Develop (x -J- h) n . 
We proceed as follows : 

u = x n ; 

du _ . 

— - = nx n ~ x \ 
ax 

g = »(» - 1) (• - 3)*-*; 

etc. etc. 

By substitution in the general formula we find 

(x + ft) n = * n + j z"- 1 h + n( \7 % ^ x "~ 2 h 

T T72 T 3 * A -r • ' ■ 

2. Develop the exponential function a x + h in powers of A. 

^m. *"(l + log a\ + (log a)'^ +...). 

3. sin (a: + ^)- 4« cos (* -h A). 

5. sin (x — ft). 6. cos (re — A). 

7. log (3 + A)- 8. log (3 - fy 

x -f ft 

9. log 7. 10. log COS 3. 

11. cos 2 (:c + ft). 12. sin 2 (a; — ft). 

13. tan<-*> (x + ft). 14. sm'-^aj-ft). 

15. Deduce the general formula 



Ar 



=/W-rf^<"»+(TT^T¥-«" 



16. Prove, by differentiation and applying the algebraic 
theorem that in two equal series the coefficients of like 
powers of the variables must be equal, that if we have 

log (a + a x x + a,x* + ...) = b + ft x s +£ 3 z 2 + ..., 



DEVELOPMENTS IN SERIES. Ill 

then the coefficients a and b are connected by the relations 

«A = «i; 

2a J 3 -f- « 1 5 1 = 2a 2 ; 
3a £ 3 + 2a A + «A - 3a 3 ; 
etc. etc. etc. 

1 7. Hence show that is the logarithm of the sum of 

J. — X 

an infinite series whose first terms are 

L . . 3x* . 13a; 3 . 73a; 4 , \ 

r + * + Tt-nr+-s-+v-> 

63. Identity of Taylor's and Maclaurin's Theorems. 
These two theorems, though different in form, are identical 
in principle. 

To see how Taylor's theorem flows from Maclaurin's, notice 
that h in the former corresponds to x in the latter. The de- 
rivatives with respect to x in Taylor's theorem are the same 
as the derivatives with respect to h } and if we suppose h = 
after differentiation Taylor's form of development can be de- 
rived at once from Maclaurin's. 

Conversely , Maclaurin's theorem may be regarded as a 
special case of Taylor's theorem, in which we take zero as the 
original value of the variable, and thus make the increment 
equal to the variable. That is, if we put f{x) in the form 

/(0 + *), 

and then, using x for li, develop in powers of x by Taylor's 
theorem, we shall have Maclaurin's theorem. 

64. Cases of Failure of Taylor's and Maclaurin's 
Theorems. In order that a development in powers of a vari- 
ble may have a determinate value it is necessary that none of 
the coefficients in the development shall become infinite and 
that the developed series shall be convergent. 

For example, cosec x cannot be developed in powers of x } 
because when x = the cosecant and all its derivatives be- 
come infinite. 



112 TEE DIFFERENTIAL CALCULUS. 

65. Extension of Taylor's Theorem to Functions of Several 
Variables, Let us have the function 

u=f{x,y). (l) 

It is required to develop this function when x and y both re- 
ceive increments. 

Let us first assign to x the increment li, and suppose y to 
remain constant. We then have,, by Taylor's theorem, 

n . du h c?u ¥ cPu h s 

f( X + h,y)= U + — r+ ^ Wl+ - ?il + m ., $ (2) 

in which u, -=—> etc., are all functions of y. 

Next, assign to y the increment k. The first member of 
(2) will become f(x -\- li, y -j- k). Developing the coefficients 
in the second member in powers of k 3 the result will be: 

u will be changed into 

du h d 2 u k % d*u k 3 
U + ty T + dtf 2 ! + dy~ 3 3 ! + ' " * ; 

-T- = D x u will be changed into 

~ , d'Djuk , d*D x uk* , 

d 3 u 

j-i = A, a w will be changed into 

na .drD^uk .dTD^iik 2 
^ W + ~^~l + ~^~2!+*°* ' 
etc. etc. etc. 

Substituting these changed values of the coefficients in (2) 

it will become 

-, . T . , x . du k , d 2 w F , d 8 w & 3 , 

t/w A d 2 w 7i & rf 3 w 7i & s 
+ ^c" I + da% ir dx~ay 1 2 ! + ' * * 

£?a J d 3 u h'k d*u If k* 
+ dx* 2! + dx*dy2l 1 + ^ 2 ^ 2 2! 2! 
,d*utf 
+ rfz 3 3!+ 



DEVELOPMENTS IN SERIES. 113 

Thus the function is developed in powers and products of 
the increments h and h. 

The law of the series will be seen most clearly by using the 
Z)-notation. For each pair of positive and integral values of 
m and n we shall have the term 

D x m D y n u X — -.*-> 

If we collect in one line the terms of the development 
which are of the same order in li and h, we shall have: 

Order of 

Terms ' h k 

1st. D x u- + D v u~. 

2d. D^ + D^D^^ + DJu^. 

3d. ZV«gj + D x *D v n |j \ + DJD/u \ |j + W* gj; 



rth. W4l + ^ r_1 ^F=l)l T + 



EXERCISES. 

i. Show that in the preceding development the terms of 
the rth order may be written in the form 



i | hrj) x r u + ^h r - l kD x r - ^D v u + f^jh r- WD x r ~ *D v *u + . . . L 
=-], f h"L etc., denoting the binomial coefficients as in § 5. 



2. Extend the development to the case of three independent 
variables, and show that the terms to the second order in- 
clusive will be as follows : 



114 

If 



THE DIFFERENTIAL CALCULUS. 



then /* (x -f- h, y -f Jc, z + I) = w 

+ D x wh + DyU'h -f Z\w J 

+ D x iiD z u • hi + DyitDgU • hi. 

66. Hyperbolic Functions. The sine and cosine of an 
imaginary arc may be found as follows: In the developments 
for sin z and cos #, namely, 



sm x 



cos 2 



3! + 5! 



2! 



2f 

4! 



let us put yi for z. (i=V—l). We thus have 
Bin |r.- = «-(jr + !j + fj + :..);] 



.• _ i _i_ y 



cos yx = 1 + |j + 1, + . . 



a) 



We conclude: 

^TAe cosine of a purely imaginary arc is real and greater 
than unity, ivhile its sine is purely imaginary. 
We find from (1), 



cos yi + i sin yi = 1 



y' 



etc. = e~ v \ 



cos yi — i sin yt = 1 + y + |-j + e ^c = e y ; 

and, by addition and subtraction, 

cos yi = i(e~ y + e y ); 

i sin y* = i(e~ v — e y ); 

sin yi = \i(e y — e _y ). 

The cosine of yi is called the hyperbolic cosine of y, 

and is written cosh y, the letter h meaning "hyperbolic." 



DEVELOPMENTS IN SERIES. 115 

The real factor in the sine of yi is called the hyperbolic 
sine of y, and is written sinh y. 

Thus the hyperbolic sine and cosine of a real quantity are 
real functions defined by the equations 



sinh y — \{f — 



:".!:( 



cosh y = i(e y -f e~ v ). 

By analogy, we introduce the additional function 

qv G - y 

tanh y = — — . 

s e y + e~ y 

The differentiation of these expressions gives 

d sinh y . d cosh y . , /rt . 

-^--•- = cosh y; __-* =sm hy; (2) 

d tanh y = — ~r -• 
J cosh y 

They also give the relations 

cosh 2 y — sinh 2 y = 1. (3) 

Inverse Hyperbolic Functions. When we form the inverse 
function, we may put 

u = cosh y. 

Then, solving the equation 

e v_|_e~» — 2 cosh y = 2u, 



we find e y = u± Vu % — 1. 

Hence 

y = log (w ± VV — 1) = cosh <-» «L (4) 

In the same way, if we put 

u = sinh y, 
we find 



y = log (w ± Vu* + 1) = sinh'- 1 * w. (5) 

From the equations (2) and (3) we find, for the derivatives 
of the inverse functions: 



cosh (_ 1} u, 


or 


u — 


: COSh 


y> 


dy_ _ 


1 








du 


♦V- 


-1 




sinh (-1) u, 


or 


u = 


sinh 


y> 


dy 


1 


• 







116 THE DIFFERENTIAL CALCULUS. 

When y 

then 

When \ 



then -f- = - (7) 

Remark. The above functions are called hyperbolic be- 
cause sinh y and cosh y may be represented by the co-ordinates 
of points on an equilateral hyperbola whose semi-axis is unity. 
The equation of such an hyperbola is 

z 2 - f = 1, 
which is of the same form as (3). 

EXERCISES. 

i. By continuing the differentiation begun in (2) prove 
the following equations: 

DJ si nn x = smh. x; 
D x * cosh x = cosh x; 
D x n ~ 2 sinh x = sinh x. 
etc. etc. 

2. Develop sinh x, as denned in (1), in powers of x byMac- 
laurin's theorem. 

v 3 y b 

Ans. sinh x = y + | j -f ^ + . . . . 

3. Develop sinh (a; -f- A) and cosh (x -|- A) by Taylor's 
theorem and deduce 

sinh (x+h) — sinh x(l + - } + . . . 1+ cosh a; fa? -j-- f -f- . . . J 

= sinh x cosh 7z. -f- cosh x sinh A; 
cosh (x-\-h) = cosh z cosh h ~f- sinh a; sinh h. 



MAXIMA AND MINIMA. 117 



CHAPTER IX. 

MAXIMA AND MINIMA OF FUNCTIONS OF A 
SINGLE VARIABLE. 

67. Def. A maximum value of a function is one which 
is greater than the values immediately preceding and follow- 
ing it. 

A minimum value is one which is less than tie values 
immediately preceding and following it. 

Remaek. Since a maximum or minimum value does not 
mean the greatest or least possible value, a function may 
have several maxima or minima. 

68. Peoblem. Having given a function 

V = <f>(*)> 

it is required to find those values of x for which y is a maxi- 
mum or a minimum. 

Let us assign to x the increments -j- h and — h, and develop 
in powers of h. We shall then have 

y' = <H* - A) = y - J^ + |f A _ etc.; 

„ ,, . , x i dy h . d*y h* . , 

/'= <H*+ h) = y + JL _ + _* _ + etc. 

In order that the value of y = <p{x) may be a minimum, it 
must, however small we suppose h, be less than either y' or 
y". That is, the expressions 

, dy 1i . cTy h* 

* * dx 1 ' dx*l'% ' 

„ , dy h d^y h* . , 



118 THE DIFFERENTIAL CALCULUS. 



must both be positive as h approaches zero. But if -~ is 

finite, li may always be made so small that the terms in A a 
shall be less in absolute magnitude than those in h (§ 14), and 
the condition of a minimum cannot be satisfied. We must 
therefore have, as the first condition, 

S= «"<*> = «■ w 

By solving this equation with respect to x will be found a 
value of x called a critical value. 

The same reasoning applies to the case of a maximum, so 
that the condition (1) is necessary to either a maximum or a 
minimum. Supposing it fulfilled, we have 

dhi h 2 d s y h* , L 

y - y = -& Fa-.E»F»^ + etc - ; 

„ d*y ¥ , d 3 y h 3 . , 

y -i = d^m + di'v¥3+ et0 - 

Since h* is positive, the algebraic sign of these quantities, 

d*y 
as h approaches zero, will be the same as that of -j\. 

When this second derivative is positive for the critical value 
of x, y, being less than y' or y" , will be a minimum. 

When negative, y will be greater than either y f or y", and 
so will be a maximum. 



We therefore conclude: 



d V _n. #y 



Conditions of minimum: -^- = 0; y^ positive. 

Conditions of maximum: -j- = 0; -~ negative. 

ax ax 

We have, therefore, the rule: 

Equate the first derivative of the function to zero. This 
equation will give one or more values of the independent vari- 
able, called critical values, and thence corresponding values of 
the function* 



MAXIMA AND MINIMA. 



119 



Substitute the critical values in the expression for the second 
derivative. When the result is positive, the function is a 
minimum; when negative, a maximum. 

Exceptional Cases. It may happen that the second deriva- 
tive is zero for a critical value of x. We shall then have 



y -y 



y" - y = 



d 3 y h 3 d*y h 4 

dx 1 3! + ^? 4 ! 



dx 3 3l 



dx* 4! 



etc.; 
etc.; 



and there can be neither a maximum nor a minimum unless 
d s y 



dx 3 



= 0. If this condition is fulfilled, y will be a maximum 



when the fourth derivative is negative; a minimum when it 
is positive. 

Continuing the reasoning, we are led to the following ex- 
tension of the rule: 

Find the first derivative in order which does not vanish 
for a critical value of the independent variable. If this de- 
rivative is of an odd order, there is neither a maximum nor a 
minimum; if of an even order, there is a minimum when the 
derivative is positive, a maximum when it is negative. 

The above reasoning may be illustrated by the graphic rep- 
resentation of the function. When the ordinate of the curve 
is a maximum or a minimum the tangent will be parallel to the 
axis of abscissas, and the angle which it makes with this axis 
will change from positive 
to negative at a point hav- 
ing a maximum ordinate, 
and from negative to posi- 
tive at a point having a 
minimum ordinate. 

For example, in the fig- 
ure a minimum ordinate occurs at the point Q, and maxi- 
mum ordinates at P and R. 




Fig. 12. 



120 THE DIFFERENTIAL CALCULUS. 



EXAMPLES AND EXERCISES. 

i. Find the maximum and minimum values of the expres- 
sion 

y = 2x 5 + 3x* - 36x + 15. 

By differentiation, 

^L = 6a; 2 + 6x - 36; 
cix 

P, = \%x + 6. 

dx 

Equating the first derivative to zero, we have the quadratic 
equation 

x* + x - 6 = 0, 

of which the roots are x = 2 and x — — 3. 

d*x 

The values of — 2 are + 30 and — 30. 

Hence x = 2 gives a minimum value of y = — 29; 

x = — 3 gives a maximum value of y = -f- 95. 

Find the maximum and minimum values of the following 
functions: 

2. x i + 3z 2 — 2±x + 9. 3. £ 3 - dx + 5. 

a; ce 2 — ^ -{- 1 

4- y = 1+!? 5 - y = * + x-l' 
, logs; logo; 

8. y = of. g. y = sin 2# — #. 

io. y = (x + l)(z - 2)'. n. ^ = (a - « 2 )(a; - J) 3 . 
_ (x + 3) s _ (x-a)(x- I) 

12t y ~ (x+ 2y I3 ' y ~ (x - P )(x - q y 

14. y = cos 2x. 15. y = cos nx. 

16. y = sin 3a. 17. y = sin ra#. 

2 ^dws. A maximum when x — -fcos a;. 



>- V 



1 -j- a; tan z' A minimum when x = —cos x. 



MAXIMA AND MINIMA. 



121 



*9- y 

21. y — 



sm x cos x. 
sin z 



20. y = 
22. */ = 



sm x cos #. 
cos Z 



1 -|- tan x * 1 + tan a;" 

23. The sum of two adjacent sides of a rectangle is equal 
to a fixed line a. Into what parts must a be divided that the 
rectangle may be a maximum? Ans. Each part = \a. 

Note that the expression for the area is x(a — x). 

24. Into what parts must a number be divided in order 
that the product of one part by the square of the other may 
be a maximum? Ans. Into parts whose ratio is 1 : 2. 

Note that if a be the number, the parts may be called x and a — x. 

25. Into what two parts must a number be divided in order 
that the product of the mth power of one part into the nth 
power of the other may be a maximum? 

Ans. Into parts whose ratio is m : n, 

26. Show that the quadratic function ax* -\~ bx -f- c can have 
but one critical value, and that it 
will depend upon the sign of the 
coefficient a whether that value 
is a maximum or a minimum. 

27. A line is required to pass 
through a fixed point P, whose 
co-ordinates are a and b in the 
plane of a pair of rectangular 
axes OX and OY. What angle 
must the line make with the axis 
of X, that the area of the triangle XYO maybe a minimum? 
Show also that P must bisect the segment XY. 

Express the intercepts which the line cuts off from the axes in terms of 
a, b and the variable angle a. The half product of these intercepts will 
be the area. 

We shall thus find 




Fig. 13. 



2 Area = (a + b cot a)(b -\- a tan a) = %ab 4- a' 2 tan a 



•>- 



tan a 



122 THE DIFFERENTIAL CALCULUS. 

Then, taking tan a = t as the independent variable, we readily find, for 
the critical values of I and a> 

b 
t = ± - , or a sin a = ± b cos a. 
a 

It is then to be shown that both values of t give minima values of the 
area ; that the one minimum area is 2ab, and the other zero ; that in the 
first case the line YX is bisected at P, and in the other case passes 
through 0. 

28. Show by the preceding figure that whatever be the an- 
gle XO Y, the area of the triangle will be a minimum when 
the line turning on P is bisected at P. 

The student should do this by drawing through P a line making a 
small angle with XPY. The increment of the area XOY will then be 
the difference of the two small triangles thus formed. Then let the small 
angle become infinitesimal, and show that the increment of the area 
XOFcan become an infinitesimal of the second order only when PX = 
PY. 

29. A carpenter has boards enough for a fence 40 feet in 
length, which is to form three sides of an enclosure bounded 
on the fourth by a wall already built. What are the sides 
and area of the largest enclosure he can build out of his ma- 
terial? Ans. 10 X 20 feet = 200 square feet. 

30. A square piece of tin is to have a square cut out from 
each corner, and the four projecting flaps are to be bent up so 
as to form a vessel. What must be the side of the part cut 
out that the contents of the vessel may be a maximum? 

Ans. One sixth the side of the square. 

31. If, in this case, the tin is a rectangle whose sides are 
2a and 2b, show that the side of the flap is 

\(a + b - Va? -ab + V). 

32. What is the form of the rectan- 
gle of greatest area which can be drawn 
in a semicircle? 

Note that if r be the radius of the circle. 



and x the altitude of the rectangle, 4/r 2 
will be half the base of the rectangle. 




Fig. 14. 



MAXIMA AND MINIMA. 123 

69. Case when the function which is to be a maximum or 
minimum is expressed as a J miction of two or more variables 
connected by equations of condition. 

The function which is to be a maximum or minimum may 
be expressed as a function of two variables, x and y, thus: 

u = <p(x, y). (1) 

If x and y are independent of each other, the problem is 
different from that now treated. 

If between them there exists some relation 

f(x, y) = 0, (2) 

we may, by solving this equation, express one in terms of the 
other, say y in terms of x. Then substituting this value of 
y in (1), u will be a function of x alone, which we may treat 
as before. 

It may be, however, that the solution of the equation (2) 
will be long or troublesome. We may then avoid it by the 
method of § 41. From (1) we have 

du _ (du\ (du\dy 
dx ~~ \dx J \dy Jdx ' 

and from (2) we have, by the method of § 37, 

dy = A,/ 

dx ~ D y f 

Substituting this value in the preceding equation, we shall 

du 
have the value of -=-, which is to be equated to zero. The 

equation thus formed, combined with (2), will give the critical 
values of both x and y, and hence the maximum or minimum 
value of u 9 



124 THE DIFFERENTIAL CALCULUS. 



EXAMPLES AND EXERCISES. 

i. To find the form of that cylinder which has the maxi- 
mum volume with a given extent of surface. 

The total extent of surface includes the two ends and the convex 
cylindrical surface. If r be the radius of the base, and h the altitude, 
we shall have : 

Area of base, nr* 1 . 

Area of convex surface, 2itrTi. 



Hence total surface = 27r(r 2 -f- rh) = const. = a. 


(a) 


Also, volume = nr^h. 


<») 


Putting u for the volume, we have, from (b), 




du dh 

— - = 27trh + nr 7 —. 
dr dr 




From (a) we find — - = ! . 

7 dr r 




Whence — = itrh — 2^. 
dr 





Equating this to zero, we find that the altitude of the cylinder must be 
equal to the diameter of its base. 

2. Find the shape of the largest cylindrical tin mug which 
can be made with a given weight of tin. 

This problem differs from the preceding one in that the top is sup- 
posed to be open, so that the total surface is that of the base and con- 
vex portion. 

Ans. Altitude = radius of bottom. 

3. Find the maximum rectangle which can be inscribed in 
a given ellipse. 

If the equation of the ellipse is b' 2 x 2 -j- a 2 ?/ 2 = aW, the sides of the 
rectangle are 2x and 2y. Hence the function to be a maximum is Axy, 
subject to the condition expressed by the equation of the ellipse. This 
condition gives 

dv b^x 



MAXIMA AND MINIMA. 



125 




We shall find the rectangle to be a maximum when its sides are 
proportional to the corresponding axes of the ellipse; each side is then 
equal to the corresponding axis divided by y%. 

4. Find the maximum 
rectangle which can be 
inscribed in the segment 
of a parabola whose semi- 
parameter is p, cut off by 
a double ordinate whose 
distance, OX, from the 
vertex is a. Show also 
that the ratio of its area 
to that of the circum- 
scribed rectangle is con- 
stant and equal to 

2 : V27. 

By taking x and y as in the Fig. 15. 

figure, a — x will be the base 

of the rectangle, and we shall have 2y for its altitude. Hence its area 
will be 2y(a — x), while x and y will be connected by the equation of the 
parabola, y* = 2px. 

5. Find the cone of maximum volume which shall have a 
given extent of conical surface. 

Ans. Alt. = radius of base X V%. 

6. Find the volume of the maximum cylinder which can be 
inscribed in a given right cone, and show that the ratio of its 
volume to that of the cone is 4 : 9. 

7. Find the cylinder of maximum cylindrical surface which 
can be inscribed in a right cone. 

Ans, Alt. of cylinder = J alt. of cone. 

8. Find the maximum cone which can be inscribed in a 
given sphere. 

If we make a central section of the sphere through the vertex of the 
cone, the base and slant height of the cone will be the base and equal 



126 THE DIFFERENTIAL CALCULUS. 

sides of an isosceles triangle inscribed in the circular section. Thus the 
equation between the base and altitude of the cone can be obtained. 

Ans. Alt. = f radius of sphere. 

9. Find the maximum cylinder which can be inscribed in 

an ellipsoid of revolution. 

o 
Ans. Alt. = — - of axis of revolution. 

VI 

10. Find the cone of maximum conical surface which can 
be inscribed in a given sphere. 

11. Of all cones having the same slant height, which has 
the maximum volume ? 

12. A boatman 3 miles from the shore wishes, by rowing 
to the shore and then walking, to reach in the shortest time 
a point on the beach 5 miles from the nearest point of the 
shore. If he can pull 4 miles an hour and walk 5 miles an 
hour, to what point of the beach should he direct his course? 

Ans. 4 miles from the nearest point of the shore. 

Express the whole time required in terms of the distance x of his point 
of landing from the nearest point of the shore. 

13. Find the maximum cone which can be inscribed in a 
paraboloid of revolution, the vertex of the cone being at the 
centre of the base of the paraboloid. 

Ans. Alt. = -J alt. of paraboloid. 

14. Find the maximum cylinder which can be described in 
a paraboloid of revolution. 

15. Find the rectangle of maximum perimeter which can 
be inscribed in an ellipse. 

16. On the axis of the parabola y 2 = 2px a point is taken 
at distance a from the vertex. Find the abscissa of the near- 
est point of the curve. 

Begin by expressing the square of the distance from the fixed point to 
the variable point (x, y) on the parabola. 

17. Determine the cone of minimum volume which can be 
circumscribed around a given sphere. 



MAXIMA AND MINIMA. 127 

1 8. Determine the cone of minimum conical surface which 
can be circumscribed around a given sphere. 

19. Find that point on the line joining the centres of two 
circles from which the greatest length of the combined cir- 
cumferences will be visible. 

20. Find that point on the line joining the centres of two 

spheres of radii a and I respectively from which the greatest 

extent of spherical surface will be visible. 

3 3 
Ans. The point dividing the central line in the ratio a* : b . 

21. Show that of all circular sectors described with a given 
perimeter, that of maximum area has the arc equal to double 
the radius. 

22. A ship steaming north 12 knots an hour sights an- 
other ship 10 miles ahead, steaming east 9 knots. What will 
be the least distance between the ships if each keeps on her 
course, and at what time will it occur? 

Ans. Time, 32 min. ; distance, 6 miles. 

23. What sector must be taken from a given circle that it 
may form the curved surface of a cone of maximum volume? 

Ans. V f of the circle. 

24. A Norman window, consisting of a rectangle sur- 
mounted by a semicircle, is to admit the maximum amount 
of light with a given perimeter. Show that the base of the 
rectangle must be double its altitude. 



128 THE DIFFERENTIAL CALCULUS. 



CHAPTER X. 

INDETERMINATE FORMS. 
70. Let us consider the fraction 

For any value we may assign to x there will be a definite 
value of <t>{x) found by dividing the numerator of the frac- 
tion by the denominator. 

To this statement there is one exception, the case of x = 3. 
Assigning this value to x, we have 

0(3) = f 

"Now, the quotient of two zeros is essentially indeterminate. 
For the quotient of any two quantities is that quantity 
which, multiplied by the divisor, will produce the dividend. 
But any quantity whetever when multiplied by will pro- 
duce 0. Hence, when divisor and dividend are both zero, 
any quantity whatever may be their quotient. 

But when we consider the terms of the fraction, not as ab- 
solute zeros, but as quantities approaching zero as a limit, 
then their quotient may approach a definite limit. We then 
regard this limit as the value of the fraction corresponding 
to zero values of its terms. 

As another example, consider the quantity 

We may compute the value of this expression for any value 
of x except 2. When x = 2 the terms will both become in- 
finite. Since if any quantity whatever be added to an infinite 



INDETERMINATE EOBMS. 129 

the sum will be infinite, it follows that any quantity what- 
ever may be the difference of two infinites. 

There are several other indeterminate forms. The follow- 
ing are the principal ones which take an algebraic form: 

J; ~; X o>; oo - oo ; 0°; <x>°; 1-. 

71. Evaluation of the Form -jf. In many cases the inde- 
terminate character of an expression may be removed by 
algebraic transformation. For example, dividing both terms 
of the fraction (1) by x — 3, it becomes x -\- 3, a determinate 
quantity even for x = 3. Again, the expression (2) can be 

reduced to the form — — -. which becomes £ when x — 2. 

x-\- 2 

The general method of dealing with the first form is as 
follows: Let the given fraction be 

0(s) 

and let it be supposed that both terms of this fraction vanish 
when x — a, so that we have 

0(a) = and 0(a) = 0. (3) 

Put h=x — a, and develop the terms in powers of h by 
Taylor's theorem. We shall then have 

<P(x) = 0(a + h) = 0(a) + hcj>'(a) + ^0'» + • • . ; 

1>(x) = tp(a + h) = 0(a) + &0'(a) + ps*"(a) + . . . ; 

whence, for the value of the fraction (comp. Eq. (3)), 

m = 0g + A^>) + • • • 
* (9,) rw + ^"(«) + . . .' 

Now, when h approaches zero as a limit, the value of this 
fraction approaches 

4>'{a) 
f'(a) 



130 THE DIFFERENTIAL CALCULUS. 

as a limit, which is therefore the required limit of the frac- 
tion when both its members approach the limit zero. 

It may happen that cp'(a) and f(a) both vanish. In this 
case the required limit of the fraction in (4) is seen to be 

In general: The required limit is the ratio of the first pair 
of derivatives of like order which do not both vanish. 

If the first derivative which vanishes is not of the same 
order in the two terms, — for example, if, of the two quantities 
<p f {a) and tp'{a), one vanishes and the other does not, — then 
the limit of the fraction will be zero or infinity according as the 
vanishing derivative is that of the numerator or denominator. 

Eemaek. It often happens that the terms of the fraction 
can be developed in the form (4) without forming the succes- 
sive derivatives. It will then be simpler to use this develop- 
ment instead of forming the derivatives. 

EXAMPLES AND EXERCISES. 
V _ /> 2 

for x = a.* 



x — a 

<p(x) = x 2 — a'; <p'(x) — 2x\ . • . <p'(a) = 2a; 

tp(x) = x — a; *p'(x) = 1; .'. ip'(a) = 1- 

. ■ . lim. (x = a) = 2a, 

x — a v y 

a result readily obtained by reducing the fraction to its lowest 
terms. 



log X . 
2. 6 • for x = l. 

x — 1 


Ans, 1. 


0* g — X 

3. for x = 0. 


Ans. 2. 



* Using strictly the notation of limits, we should define the quantity 
sought as the limit of the fraction when x approaches the limit a. But 
no confusion need arise from regarding the limit of the fraction as its 
value for x = a, as is customary. 



INDETERMINATE FORMS. 131 

x — sin x 

4. - 3 for (x = 0). Ans. f 

Here the successive derivatives of the terms are: 

<p'(x) = 1 — cos x; <p"(x) = sin x; <p'"(x) = cos x. 
ib'(x) = dx*; $"\x) = 6x; f"(x) = 6. 

The third derivatives are the first ones which do not vanish 
for x = 0. 

5. for x = 0. Ans. log a— log b— log -. 

, tan x — sin x 

6. r for x — 0. -4 «s. 3. 

re — sin x 

7. ^ : for x = 0. .4 ws. -„. 

1 — cos nx n 

8. for x = l. Ans. a log a. 

x — 1 & 

a x — b x 

9. =- for x = 1. Ans. a log a —b log J. 

x — ± 

sin a: — sin a . 

IO# . . f or x = a. Ans. cos a. 

x — a 

tan y — tan « - . sec' a 

11. — r^ =— for y — a. Ans. ^—. — . 

cos 3 y — cos a a 2 sin a 

log (1 '+ x) + log (1 - a) # . . 

12. — ^-* — ■ — — 2-* '- for x = 0. .4ws. +1. 

cos re — sec x ■ 

log (^ + #) — log (a — x) , n . 2 

13. — — for re = 0. Ans. -. 

x a 

sin 2x 4- 2 sin 9 re — 2 sin x « 

14. = for a = 0. Jws. 4. 

cos re — cos x 

e x — e~ x — 2x„ 
15. : for x = 0. <4ws. 1. 

re — sin z 



_. e v 4- sin y — 1 
16. ■ = ' = •, — r- for y = 0. Ans. 2. 



1 — sin a — cos re + log (i+a;) , 
7. ^ _ j- _ z * y ^ ; for (s = 0). ^ns. 0. 



132 THE DIFFERENTIAL CALCULUS. 

72. Forms — and X oo . These forms may be reduced 

00 J 

to the preceding one by a simple transformation. Any frac- 
tion -^ may be written in the form - — : — -^ If N and D both 
D J 1 -^ N 

become infinite, 1 -^- D and 1-r J will both become infini- 
tesimal, and thus the indeterminate form of the fraction will 
be£. 

Again, if of two factors A and B, A becomes infinitesimal 
while B becomes infinite, we write the product in the form 

=- and then it is a fraction of the first form. 

1 -f- B 

But this transformation cannot always be successfully ap- 
plied unless the term which becomes infinite does so through 
having- a denominator which vanishes. For example, let it 
be required to find the limit of 

z m (log x) n 

for x = 0. Here x m approaches zero, while log x, and there- 
fore (log x) n , becomes infinite for x = 0. Hence the denomi- 
nator of the transformed fraction will be j- (putting for 

brevity I = log x). The successive derivatives of this quantity 
with respect to x are 



n n I 1 n -f- 1\ , 

+ 1 ' x\F T ~ 1 ~ ] ~T rTJ J ; 



xV 

The successive derivatives of the numerator are 

mx m ~ 1 ; m(m — l)x m ~ 2 ; etc. 
The limiting values of the given quantity x m l n thus become 
mx m l n+1 m(m — l)x m 

in + i ~r ln +») 



n 



which remain indeterminate in form how far soever we may 
carry them. 



INDETERMINATE FORMS. 133 

In such cases the required limit of the fraction can be 
found only by some device for which no general rule can be 
laid down. In the example just given the device consists in 
replacing x by a new variable y, determined by the equation 

log x = — y. 
We then have x = e ~ y . 

Since f or x ^ y = oo , we now have to find the limit of 

f or y = oo . 

By taking the successive derivatives of the two terms of 

the fraction — - we have the successive forms 

e my 

ny n ~\ n(n — l)y n ~ z m n(n — 1) (n — 2)y n ~\ 
me my ' mV* ' m 3 e my " ' 6 °* 

Whatever the value of n, we must ultimately reach an ex- 
ponent in the numerator which shall be zero or negative, and 
then the numerator will become n\ if n is a positive integer, 
and will vanish for y = oo , if n is not a positive integer. But 
the denominator will remain infinite. We therefore con- 
clude: 

lim. [x m (log x) n ] (x = 0) = 0, 

whatever be m and n, so long as m is positive. 

From this the student should show, by putting z = x" 1 and 
m — \, that the fraction 

z 

(log^r 

becomes infinite with z, how great soever the exponent n, and 
therefore that any infinite number is an infinity of higher 
order than any power of its logarithm. 

73. Form oo — oo . In this case we have an expression of 
the form 

F(x) = u — v, 



134 THE DIFFERENTIAL CALCULUS. 

in which both u and v become infinite for some value of z. 
Placing it in the form 

we see that F(x) will become infinite with u unless the fraction 

v 

— approaches unity as its limit. When this is the case the 

expression takes the form oo x Oof the preceding article. 

74. Form l 00 . To investigate this form let us find the 
limit of the expression 

when n becomes infinite. Taking the logarithm, we have 
log u = hn log (l + -J 

Making n infinite, we have 

lim. log u = h; 
or, because the limit of log u is the logarithm of lim. u, 

log lim. u = h. 

Hence lim. ( 1 -| — j (n = oo ) = e\ 

In order that this result may be finite, h itself must not be 
infinite. We therefore reach the general conclusion: 
Theorem. In order that an expression of the form 

(1 + a y 

may have a finite limit when a becomes infinitesimal and x 
infinite, the product ax must not become infinite. 

Cor. If the product ax approaches zero as a limit, the 
given expression will approach the limit unity. 



INDETERMINATE FORMS. 135 

75. Forms 0° and oo°. Let an expression taking either 
of these forms as a limit be represented by u^=F. The 
problem is to find the limiting value of the expression when 
approaches zero and u either approaches zero or becomes 
infinite. 

From the identity u = e log u 

we derive F = ift = e* loeu . 

We infer that the limit of F will depend upon that of log u. 

If lim. log u is + °° > then lim. F = <x> . 

If lim. log w is — oo , then lim. F = 0. 

If lim. log w is 0, then lim. F = 1. 

If lim. log w is finite, then lim. F is finite. 
Hence the rule: To find the limit of u^ when = and 
u — or oo , put I = lim. log u. Then 

lim. ifi = e*. 

EXAMPLES AND EXERCISES. 

i. Find lim. x x for z = 0. 
Here of — e xloex . 

Since x log z has zero as its limit when x — 0, the required 
limit is e° or 1. 

2. lim. x nx for z = 0. 

3. lim. #* for x = 00 . 

4. 2; 1-Tri for a; = 1. 

n 

5. x^* for x = 1. 

h 



6. (1 — sb)* for sc = 0. 

e x — e ~ x 
log (1 + 2 
log sin 2z 



g 3 ' g — ^ 

7. ; T-— — r for x = 0. 

log (1 + SB) 



^ws. 


F=l. 


^IftS. 


F=l. 


Ans. 


1 

e ' 


Ans. 


e~ n . 


Ans. 


e~\ 


Ans. 


2. 


Ans. 


log 2. 



for x = 0. 
log sin x 

e x + log (1 - sb) - 1 £ 

—^ — 2.-4 '- for x == 0. .4ns. £, 

x — tan sb 



136 THE DIFFERENTIAL CALCULUS. 



log x\ 

for x = 



o. (^ 



JjJS. 


1. 


^s. 


- 1. 


^4«s. 


a. 


A?is. 


log a. 


Ans. 


1. 


Ans. 


ei. 


Ans. 


e-K 


Ans. 


2 

7t 



\\. x tan x — — sec a; tor a; = — . 

12. y sm — for y = co. 

13. #(a* — ly for x = <x>. 

1 
/tan aA* 

* 4 ' \x~J f ° r * = °' 

i_ 

/tan x\x* . 

I5 ' l~"J f ° r X = ' 

j. 

16. (cos a;)** for ic = 0. 

J 7- (1 - y) tan -y for y = 1. 
1 

18 . fellY for a; = 0. Ans. 1. 

19. a; — x 2 log (l -j J for x == 00 . ^4ws. £. 

20. = — — — r- for x = 0. ^4?is. 2. 

log (1 + z) 

2 

21. (.— — - — —J for x = 0. ^ws. ^a,. 

22. — — ' — - — ■ ■ for x = 0. Ans. a,a v 

V n I * s 

23. Show that, how great soever the exponent n, 
— 00 when x = 00 . 



(log x) n 



PLANE CURVES. 137 



CHAPTER XI. 

OF PLANE CURVES. 

76. Forms of the Equations of Curves. As we have here- 
tofore considered curve lines, they have been defined by an 
equation between the co-ordinates of each point of the curve, 
and therefore of one of the forms 

y=f(x); x=f(y); (1) 

and F{x 9 y) — 0. 

The distinguishing feature of the equation is that when we 
assign a value at pleasure to one of the co-ordinates x or y, 
one or more corresponding values of the other co-ordinate are 
determined by the equation. 

But the relation between x and y may be equally well 
defined by expressing each of them as a function of an 
auxiliary variable, which is then the independent variable. 
Calling this auxiliary variable u, the equations of a curve will 
be of the form 

* = AM; I , 9 v 

y = 0,00. I w 

Assigning values at pleasure to u, we shall have correspond- 
ing values of x and y determining each point of the curve. 

An advantage of this method of representation is that for 
each value of u we have one definite point of the curve, or 
several definite points when the equations give several values 
of the co-ordinates for each value of u; and we thus have a 
relation between a point and the algebraic quantity u. 

It is also to be remarked that by eliminating u from the 
equations (2) we shall get a single equation between x and y 
which will be the equation of the curve in one of the forms 

(i). 



138 



TEE DIFFERENTIAL CALCULUS. 




Example 1. Let us put 

a, b E the co-ordinates of any fixed point B of a straight line; 

a = the angle which 
the line makes with 
the axis of x; 

p = the distance of 
any point P of the 
line from the point 
{a, I). 

Then we readily see 
from the figure that 

the co-ordinates x and Fig. 16. 

y of P are given by the equations 

x = a -j- p cos a; ' 
y = b -f- p sin #; 

which are equations of the straight line in the independent 
form. 

Here p is the auxiliary variable, called u in Eq. 2. By 
eliminating this quantity we shall have 

x sin a — y cos a = a sin a — ~b cos a, 
which is the equation of the line in one of its usual forms. 

Example 2. The equation of a circle may be expressed in 

the form 

x = a -j- c cos w, 

y = b + c sin u; 

u being the independent T 

variable. 

By writing (4) in the 

form 



(3) 



(4) 



cc — # 



c COS w, 



y — b = c sin w, 
and eliminating w by 
taking the sum of the 
squares of the two equa- 
tions, we have 




Fig. 17. 




PLANE CURVES. 139 

(x - af + {y - bf = c\ 

the equation of a circle of radius c. 

Notice the beautiful relation between (3) and (4). They 
are the same in form: if in (4) we write p for c and a for 
u, they will be the same equations. Then, by supposing p 
constant and ex variable, we are carried round the point (a, h) 
at a constant distance p, that is, around a circle. By suppos- 
ing p variable and a constant we are carried through (a, I) 
in a constant direction, that is, along a straight line. 

77. Infinitesimal Elements of Curves. Let P and P' be 
two points on a curve, P being supposed 
fixed, and P' variable. We may then sup- 
pose P' to approach P as its limit, and in- 
quire into the limits of any magnitudes 
associated with the curve. 

We may also measure the length of an 

Fin 18 

arc of the curve from an initial point G to 
a terminal point P. Then, supposing C fixed and P variable, 
PP' may be taken as an increment of the arc. 
If we put 

s = arc OP, 

we shall have 

As = arc PP'. 

Axiom. The ratio of an infinitesimal element of a curve 
to the straight line joining its extremities approaches unity as 
its limit. 

We call this proposition an axiom because a really rigorous 
demonstration does not seem possible. Its truth will appear 
by considering that if the curve has no sharp turns, which 
we presuppose, then it can change its direction only by an in- 
finitesimal quantity in any infinitesimal portion of its length. 
Now, a line which has the same direction throughout its length 
is a straight line. 



140 



TEE DIFFERENTIAL CALCULUS. 




Fig. 19. 



78. Theorem I. If a straight line touch a curve at the 
poi?it P, a point P' on the 
curve at an infinitesimal 
distance will, in general, 
be distant from the tangent 
by an infinitesimal of the 
second order. 

Let y = f (x) be the _o 
equation of the curve. 

Let us transform the equation to a new system of co-ordi- 
nates, x' and y', so taken that the axis of X' shall be parallel 

dy' 
to the tangent at P. This will make —-, = 0. Let x' and y' 

be the co-ordinates of P, and {x' -\- h, y") the co-ordinates 
of a point P' near P. 

Developing by Taylor's theorem, we have 

y ~ y ~ lx ftl + dx* F2 + 

Now, y" — y' is the distance P'Q of the point P' from the 

dy' 
tangent at P. Since -i~ = 0, when h becomes infinitesimal 
ax 

d*y' h 2 
the term of highest order in this distance is — y 2 - =-— , a quan- 
tity of the second order. 



d\f 



Remark. In the special case when -=~ = 0, the distance 

ax 

in question may be a quantity of the third or of some higher 
order, according to the order of the first differential coeffi- 
cient which does not vanish. 

Corollary. The cosine of an infinitesimal arc differs 
from unity by an infinitesimal of the second order. 

For if we draw a unit circle with its tangent at the initial 
point, the cosine of an arc will differ from unity by the dis- 
tance from the end of the arc to the tangent line. When the 
arc is infinitesimal, the corollary follows from the theorem. 



PLANE CURVES. 141 

Theorem II. The area included between an infinitesimal 
arc and its chord is not greater than an infinitesimal of the 
third order. 

From Th. I. we may readily see that the maximum distance 
between the chord and its arc is a quantity of the second 
order. The area is less than the product of this distance by 
the length of the chord, which product is an infinitesimal of 
at least the third order. 

*79. Expressions for Elements of Curves. Defi An 
element of a geometric magnitude is an infinitesimal por- 
tion of that magnitude. 

The word implies that we conceive the magnitude to be 
made up of infinitesimal parts. 

Element of an Arc. Let us put 

s = the length of any arc of a curve; 
ds = an element of this arc. 

If P and P' be two points of a curve, we shall have 

(chord PP'Y = Ax 2 + Ay\ 

When PP f becomes infinitesimal, As 

the ratio of ds to PP' becomes unity 
(§ 77), and we have s^ & x 

ds 2 = dz* + df; Fig. 20. 




w=Vi+(th- 



Case of Polar Co-ordinates. To express the element of a 
curve referred to polar co-ordinates, differentiate the equa- 
tions 

x — r cos 6; y = r sin 0. 
Thus dz — cos ddr — r sin Odd; 

dy = sin ddr + r cos Odd; 
which gives ds* = dr* + rW 



and ds = fr J + {^f d0 ' 



142 



THE DIFFERENTIAL CALCULUS. 



80. Equations of certain Noteworthy Curves. The Cycloid. 
The cycloid is a curve described by a point on the circumfer- 
ence of a circle rolling on a straight line. A point on the 
circumference of a carriage-wheel, as the carriage moves, 
describes a series of cycloids, one for each revolution of the 
wheel. 

To find the equation of the cycloid, let P be the generating 
point. Let us take the line on which the circle rolls as the 
axis of X, and let us place the origin at the point where P 
is in contact with the line OX. 



T 








M 


P 


\a 




SI 






A 


^ 


C ] >y 


■\ 


/ \ 




a J \ / 



o 



Q R B 

Fig. 21. 

Also put 

a = the radius of the circle ; 

u = the angle through which the circle has rolled, expressed 
in terms of unit-radius. 

Then, when the circle has rolled through any distance OR, 
this distance will be equal to the length of the arc PR of the 
circle between P and the point of contact R, that is, to au. 
We thus have, for the co-ordinates of the centre, C, of the 
circle, 

x = au; 
y = a; 

and for the co-ordinates of the point P on the cycloid, 

x = au — a sin u = a(u — sin u); 
y = a —a cos u = ail — cos u)\ 

which are the equations of the cycloid with u as an independ- 
ent variable. 



.;! 



(i) 



PLANE CURVES. 143 

To eliminate u, find its value from the second equation, 

^= C os ( - i) fi ~^y 

m,. . . ./z — V2ay — y* 

This gives sin u = V 1 — cos u = — ♦ 

a 

Then, by substituting in the first equation 

x = a cos«- x > °—^- - V2ay-y% (2) 

which is the equation of the cycloid in the usual form. 

8 1 . The Lemniscate is the locus of a point, the product of 
whose distances from two fixed points (called foci) is equal 
to the square of half the distance between the foci. 

Let us take the line joining the foci as the axis of X, and 
the middle point of the segment between the foci as the 
origin. Let us also put cEhalf the distance between the 
foci. 




Fig. 22. 

Then the distances of any point {x, y) of the curve from 
the foci are 



V(x - cf + y* and V(x + cf + y\ 

Equating the product of these distances to &, squaring and 
reducing, we find 

( X ' + t/Y = 2c\x' - f), (3) 

which is the equation of the lemniscate. 



144 



THE DIFFERENTIAL CALCULUS. 



Transforming to polar co-ordinates by the substitutions 
x = r cos 0, 
y = r sin 8, 
we find, for the polar equation of the lemniscate, 

r a = 2c a cos 20. (4) 

Putting y = 0, we find, for the point in which the curve 
cuts the line joining the foci, 

x = ± V'Zc = a. 
The line a is the semi-axis of the lemniscate. Substitut- 
ing it instead of c, the rectangular and polar equations of the 
curve will become 



cos 2(9. 



(5) 



82. The Archimedean Spiral. This curve is generated 
by the uniform motion of a point along a line revolving uni- 
formly about a fixed point. 

To find its polar equation, let us take the fixed point as the 
pole, and the position of the revolving line when the generat- 
ing point leaves the pole 
as the axis of reference. 
Let us also put 

a = the distance by 
which the generating 
point moves along the 
radius vector while the 
latter is turning through 
the unit radius. 

Then, when the ra- 
dius vector has turned 
through the angle 6, the 
point will have moved 
from the pole through the distance ad. 

r = a6 




Fig. 23. 

Hence we shall have 



as the polar equation of the Archimedean spiral. 



PLANE CURVES. 145 

If we increase 6 by an entire revolution (27r), the corre- 
sponding increment of r will be 2na 9 a constant. Hence: 

The Archimedean spiral cuts any fixed position of the ra- 
dius vector in an indefinite series of ' equidist 'ant points, 

83. The Logarith7iiic Spiral, This is a spiral in which 
the logarithm of the radius vector is proportional to the angle 
through which the radius vector has moved from an initial 
position. Hence, if we put O 
for the initial angle, we have 

log r = 1(6 - 0.), 
I being a constant. Hence 

r = e = e °e . 

Putting, for brevity, 

the equation of the logarith- 
mic spiral becomes fig. 24. 

r = ae l 
a and I being constants. 




* 



EXERCISES. 

i. Show (1) that the maximum ordinate of the lemniscate 
is \c, and (2) that the circle whose diameter is the line join- 
ing the foci cuts the lemniscate at the points whose ordinates 
are a maximum. 

2. Find the following expression for the square of the dis- 
tance of a point of a cycloid from the starting point ( 0, Fig. 

21): 

r a = 2ay -f- 2uax — a*u % . 

3. A wheel makes one revolution a second around a fixed 
axis, and an insect on one of the spokes crawls from the cen- 
tre toward the circumference at the rate of one inch a second. 
Find the equation of the spiral along which he is carried. 



146 THE DIFFERENTIAL CALCULUS. 

4. If, in that logarithmic spiral for which a = 1 and 1=1, 

r = e e s 

the radius vector turns through an arc equal to log 2, its 
length will be doubled. 

5. If, in any logarithmic spiral, one radius vector bisects 
the angle between two others, show that it is a mean propor- 
tional between them. 

6. Show that the pair of equations 

x = au*, 
y = bu, 

represent a parabola whose parameter is — . 

7. If, in the equation' of the Archimedean spiral, 6 and 
therefore r take all negative values, show that we shall 
have another Archimedean spiral intersecting the spiral given 
by positive values of 6 in a series of points lying on a line at 
right angles to the initial position of the revolving line. 

This should be done in two ways. Firstly, by drawing the continua- 
tion of the spiral when, by a negative rotation of the revolving line, the 
generating point passes through the pole. It will then be seen that the 
combination of the two spirals is symmetrical with respect to the vertical 
axis. Secondly, by expressing the rectangular co-ordinates of a point of 
the spiral in terms of we have 

x = a6 cos 0, 

y = aO sin 0. 

Changing the sign of in this equation will change the sign of x and 
leave y unchanged. 

8. Show that if we draw two lines through the centre of a 
lemniscate making angles of 45° with the axes, no point of 
the curve will be contained between these lines and the axis 
of Y. 



TANGENTS AND NORMALS. 



147 



CHAPTER XII. 
TANGENTS AND NORMALS. 

84, A tangent to a curve is a straight line through two 
coincident points of the curve. 




Fig. 25. 

A normal is a straight line through a point of the curve 
perpendicular to the tangent at that point. 

The subtangent is the projection, TQ, upon the axis of 
X, of that segment TP of the tangent contained between 
the point of contact and the axis of X. 

The subnormal is the corresponding projection, QN y of 
the segment PN of the normal. 

Notice that a tangent and a normal are lines of indefinite 
length, while the subtangent and subnormal are segments of 
the axis of abscissas. Hence the former are determined by 
their equations, which will be of the first degree in x and y, 
while the latter are determined by algebraic expressions for 
their length. 

But the segments TP and PN are sometimes taken as 
lengths of the tangent and normal respectively, when we con- 
sider these lines as segments. 



148 THE DIFFERENTIAL CALCULUS. 

85. General Equation for a Tangent. The general prob- 
lem of tangents to a curve may be stated thus: 

To find the condition which the parameters of a straight 
line must satisfy in order that the line may be tangent to a 
given curve. 

But it is commonly considered in the more restricted form: 
To find the equation of a tangent to a curve at a given point on 
the curve. 

Let (x lf y 2 ) be the given point on the curve. By Analytic 
Geometry the equation of any straight line through this point 
may be expressed in the form 

y-y 1 = m(x- x x ); (5) 

m being the tangent of the angle which the line makes with 
the axis of X. But we have shown (§ 20) that 

this differential coefficient being formed by differentiating the 
equation of the curve. Hence 

y-y, = %y-^) (6) 

is the equation of the tangent to any curve at a point (x l9 yj 
on the curve. 

Equation of the Normal. The normal at the point (x l9 yj 
passes through this point, and is perpendicular to the tangent. 
If m f be its slope,, the condition that it shall be perpendicular 
to the tangent is (An. Geom.) 

m' = - — = - — 

m m & 

dx x 

Hence the equation of the normal at the point (x l9 yj is 

§^(</-y,) = *,-*• (7) 



TANGENTS AND NORMALS. 



149 



In these equations of the tangent and normal it is necessary 
to distinguish between the cases in which the symbols x and 
y represent the co-ordinates of points on the tangent or nor- 
mal line, and those where they represent the given point of 
the curve. Where both enter into the same equation, one set, 
that pertaining to the curve, must be marked by suffixes or 
accents. 

86, SuUangent and Subnormal, To find the length of 
the subtangent and subnormal, we have to find the abscissa 
x of the point T in which the tangent cuts the axis of abscis- 
sas. We then have, by definition, 




Fig. 26. 

Subtangent = x 1 — x 

The value of x is found by putting y = and x — x in 
the equation of the tangent. Thus, (6) gives 

- *• = Iy° - *»>• 

Hence, for the length of the subtangent TQ, 



Subtangent = x, — x n = -i~. 

dx x 
We find in the same way from (7), for QN, 

Subnormal = — v,-^. 
^'dx. 



(8) 



(9) 



150 



THE DIFFERENTIAL CALCULUS. 



87. Modified Forms of the Equation. In the preceding 
discussion it is assumed that the equation of the curve is given 
in the form 

y =/(*)• 

But, firstly, it may be given in the form 

F{x, y) = 0. 

We shall then have (§ 37) 

dF 
dy J _ dx 1 
dx,~ ~~ dF' 

dy, 

Substituting this value in the equations (6) and (7), we find 

(10) 



„ , dF. , dF, . 

Tangent: _(y _ y.) =_(*,_ *) 



w i dF t ^ dF i \ 

Normal: — (y _ y.) = _ (* _ *,). 

Secondly, if the curve is defined by two equations of the 
form 

0i (*), 



we have 



x= (p^u), ) 
V = 0», ) 

dy_ 
dy t _ du 
dx t dx ' 

du 



(ii) 



in which there is no need of suffixes to x and y in the second 
member, because this member is a function of u, which does 
not contain x or y. 

By substitution in (6) and (7), we find 



Eq. of tangent: (y - y^ = (x - x^. 
Eq. of normal: (y - yjg- = (*, - x) g. 



(12) 



TANGENTS AND NORMALS. 151 

CL 05 du 

By substituting in these equations for x t y 1} -j- and -j- 

their values in terms of u, the parameters of the lines will be 
functions of u. Then, for each value we assign to u, (11) 
will give the co-ordinates of a point on the curve, and (12) 
will determine the tangent and normal at that point. 

88, Tangents and Normals to the Conic Sections, Writing 
the equation of the ellipse in the form 

ay + 5 V = a*l\ (a) 

we readily find, by differentiation, 

dy _ b*x 
dx a*y' 

Applying the suffix to x and y, to show that they represent 
co-ordinates of points on the ellipse, substituting in (6) and 
(7), and noting that x x and y x satisfy (a), we readily find: 

For the tangent: 3J + M = i. 
a o 

a* b* 
For the normal: —x y = a* — b*. 

s, y? 

Taking the equation of the hyperbola, 
- ay + b*x* = a*l\ 
we find, in the same way, 

For the tangent: ^ - %r = *• 
a o 

a* b* 
For the normal: —x A — y = a* 4- b*, 

s, y: 

Taking the equation of the parabola, 

y a = %px, 
we find, by a similar process, 

For the tangent: y x y = p(x + xj. 
For the normal: y — y t = — — (», — x). 



152 THE DIFFERENTIAL CALCULUS. 

89. Problem. To find the length of the perpendicular 
dropped from the origin upon a tangent or normal. 

It is shown in Analytic Geometry that if the equation of a 
straight line be reduced to the form 

Ax+By + (7=0, 
the perpendicular upon the line from the origin is 

C 
p = — - 

VA* + B 7 

It must be noted that in the above form the symbol G rep- 
resents the sum of all the terms of the equation of the line 
which do not contain either x or y. 

If we have the equation of the line in the form 

y-y 1 = m(x- zj, 
we write it mx — y — mx x + y x = 0, 

and then we have 

A = m; 

£=-1; 

C = y x — mx x . 

Thus, the expression for the perpendicular is 

y, — mx, 
Vm* + 1 
Substituting for m the values already found for the tan- 
gent and normal respectively, we find, 
For the perpendicular on the tangent : 

1 dx y V,dx, — x,dy, _. 



^m 



For the perpendicular on the normal : 

x 4-y^ 
= l ^ Ul dx x = x x dx x + y x dy x . 



TANGENTS AND NORMALS. 



153 




Fig. 27. 



90. Tangent and Normal in Polar Co-ordinates. 

Problem. To find the 
angle which the tangent at 
any point makes with the 
radius vector of that point. 

Let PP' be a small arc 
of a curve referred to polar 
co-ordinates; 

KP, a small part of the 
radius vector of the point 
P (the pole being too far 
to the left to be shown in 
the figure); 

K'P' , the same for the point P'. 

KSR, a parallel to the axis of reference. Drop PQLK' P' . 

Let SPTbe the tangent at P. We also put 
y = angle KPS which the tangent makes with the radius 
vector. 

Then let P' approach P as its limit. Then 

QP' = dr; PQ = rd6; 
PQ . rdd 

We also have 

1 _ 1 dr 

d~0' 



(1) 



cos y 



|/(1 + tan 2 y) 



/war 



sin y = cos y tan y = 



•wen 



(2) 



Cor. The angle ESP which the tangent makes with the 
axis of reference is y -\- 6, 



154 THE DIFFERENTIAL CALCULUS. 

91. Perpendicular from the Pole upon the Tangent and 
Normal. When y is the angle between the tangent and the 
radius vector, we readily find, by geometrical construction, 
that the perpendicular from the pole upon the tangent and 
normal are, respectively, 

p = r sin y and p = r cos y. 

Substituting for sin y and cos y the values already found, 
we have, 

For the perpendicular on tangent: 

For the perpendicular on normal : 

r dr 



•war 



(3) 



92. Problem. To find the equation of the tangent and 
normal at a given point of a curve whose equation is expressed 
in polar co-ordinates. 

It is shown in Analytic Geometry that if we put 

p = the perpendicular dropped from the origin upon a line; 
a = the angle which this perpendicular makes with the 
axis of X; 

the equation of the line may be written 

x cos a -j- y sin a — p = 0. (1) 

Now, as just shown, the tangent makes the angle y -f- 6 
with the axis of X, and the perpendicular dropped upon it 
makes an angle 90° less than this. Hence we have 

a = y + - 90°; 
cos a — sin (y -J- 6) = sin y cos 6 -f- cos y sin 6; 
sin a = — cos (y -f- &) = — cos y cos 6 -\- sin y sin 6. 



TANGENTS AND NOBMALS. 155 

By substitution in (1), the equation of the tangent becomes 

a;(sin y cos 6 + cos y sin 6) 

— y(cos y cos 6 — sin y sin 6) — p — 0. 

Substituting for cos y, sin y and p the values already found, 
this equation of the tangent reduces to 

(r cos -\~ -Th sni 0\x -\-{r sin — -^ cos #1 # — r 3 = 0, (2) 

r and being the co-ordinates of the point of tangency. 

In the case of the normal the perpendicular upon it is 
parallel to the tangent. Therefore, to find the equation of 
the normal, we must put in (1) 

a = y -\- 6. 

Substituting this value of a, and proceeding as in the case 
of the tangent, we find, for the normal, 

\—.-Q cos 6 — r sin 6 J x + ( r cos 6 -f- -j* sin 6) y — r-^- a — 0. (3) 

Generally these equations will be more convenient in use if 
we divide them throughout by r. Thus we have: 

Equation of the tangent : 

(cos 6 + - — sin 6 J x + (sin B — --~ cos 6jy-r = 0. (4) 

Equation of the normal : 

&i cos e - sin e ) x + &S sin " + cos e )y - % = °- < 5) 

In using these equations it must be noticed that the co- 
efficients of x and y are functions of r and 6, the polar co- 
ordinates of the point of tangency. When r 3 6 and -j- a are 

given, this point and the tangent through it are completely 
determined, 



156 THE DIFFERENTIAL CALCULUS. 

EXERCISES. 

i. Show that in the case of the Archimedean spiral the 
general expressions for the perpendiculars from the pole upon 
the tangent and normal, respectively, are 

ad' ■ aO 

*= */(l + 3 ) ^ ^ = 1/0 + 0? 

Thence define at what point of the spiral the radius vector 
makes angles of 45° with the tangent and normal. Find also 
what limit the perpendicular upon the normal approaches 
as the folds of the spiral are continued out to infinity. 

Show also from § 92 that the tangent is perpendicular to 
the line of reference at every point for which 

r sin 6 — a cos 6 = 0, 

and hence that, as the folds of the spiral are traced out to 
infinity, the ordinates of the points of contact of such a tan- 
gent approach ± a as their limit. 

2. Show by Eq. 12 that in the case of the logarithmic 
spiral the angle which the radius vector makes with the tan- 
gent is a constant, given by the equation 

tan y = —. 

3. Show from Eq. 12 that if a curve passes through the 

pole, the tangent at that point coincides with the radius 

dr 
vector, unless — = at this point. Thence show that in the 

lemniscate the tangents at the origin each cut the axes at 
angles of 45°. 

4. Show that the double area of the triangle formed by a 

tangent to an ellipse and its axes is . Then show that the 

X \V\ 

area is a maximum when — = ± q ~. 

a 

Show also" that the area of the triangle formed by a nor- 
mal and the axes is a maximum for the same point. 



ASYMPTOTES AND SINGULAR POINTS. 157 



CHAPTER XIII. 

OF ASYMPTOTES, SINGULAR POINTS AND 
CURVE-TRACING. 

93. Asymptotes. An asymptote of a curve is the limit 
which the tangent approaches when the point of contact re- 
cedes to infinity. 

In order that a curve may have a real asymptote, it must 
extend to infinity, and the perpendicular from the origin upon 
the tangent must then approach a finite limit. 

For the first condition it suffices to show that to an infi- 
nite value of one co-ordinate corresponds a real value, finite 
or infinite, of the other. 

For the second condition it suffices to show that the expres- 
sion for the perpendicular upon the tangent (§§89, 91) ap- 
proaches a finite limit when one co-ordinate of the point of 
contact becomes infinite. If, as will generally be most con- 
venient, the equation of the curve is written in the form 

F{x, y) = 0, (1) 

the value (1) of the perpendicular, omitting suffixes, may be 
reduced to 

dF , dF 
y dti +X dx~ 



S(§f)'+(f}y 



\dy 

If this expression approaches a real finite limit for an 
infinite value of x or y, the curve has an asymptote. 

If the curve is referred to polar co-ordinates, we use the 
expression (3), § 91, for p. If this approaches a real finite 
limit for an infinite value of r, the curve has an asymptote. 



158 



THE DIFFERENTIAL CALCULUS. 



The existence of the asymptote being thus established, its 
equation may generally be found from the form (10), § 87, 
which we may write thus: 



dF . dF 



dF 



dF 



dx x dy x l dx x dy x 



(3) 



by supposing x x or y x to become infinite. 

dF dF 

Commonly the coefficients -r- and -=— will themselves be- 
J dx x dy x 

come infinite with the co-ordinates. We must then divide 

the whole equation by such powers of x x and y x that none of 

the terms shall become infinite. 



94. Examples of Asymptotes, 

1. F(x) — x z + y 3 — Saxy = 0.(a) 

The curve represented by this 
equation is called the Folium of 
Descartes. The equation (3) gives 
in this case, applying suffixes, 

« - ayd x + to i* ~ ax i)y 

= x ? + V? - %**& = WiVi- 
To make the coefficients of x and 
y finite for x x = <x> , divide by x x y x . 
comes 









Fig. 28. 



Then the equation be- 



6-l)*+(H>-=»- » 

Let us now find from (a) the limit of y x for 2^ = 00. We 



l + fS = 3„fv 



The second member of this equation will approach zero as 
a limit, unless y x is an infinite of as high an order as x x , 
which is impossible, because then the first member of the 
equation containing y* would be an infinite of higher order 



ASYMPTOTES AND SINGULAR POINTS. 



159 



than the second member, which is absurd. Hence, passing 
to the limit, 

Urn. (£j (*, = «,) = - 1. 

Then, by substitution in (b), we find, for the asymptote, 

z + y + a = o. 

2. Take next the equation 

p(x, y) = x 3 — %x i y — ax 2 — a*y = 0. (a) 

With this equation (3) becomes 

(3a;/ - 4=x 1 y 1 - 2axjx - (2x 1 i + a*)y 

= Sx? - 6x 1 *y 1 - 2ax* - a*y x . (b) 




Fig. 29. 

We notice that the terms of highest order in the second 

member are three times those of highest order in (a). From 

(a) we have 

x* - 2x*y = ax* + a*y x . 

Substituting in the second member of (b), and dividing by 
z* t (b) becomes 



( 3 - 4 I;-!)-K|>-+^- *' 



Solving (a) for y, we find 

Vx _ <' 



ax. 



x 1 2x* + a* 9 
an expression which approaches the limit $ when x 1 = oo . 
Thus, passing to the limit, (b') gives, for the equation of the 

asymptote, 

x •— 2y = a. 




160 THE DIFFERENTIAL CALCULUS. 

3. The Witch of Agnesi. This curve is named after the 
Italian lady who first investigated its 
properties. Its equation is 

x*y -f- a?y — a 3 = 0. (a) 

The equation of the tangent is 

%x Y y x x + (x? + ajy = Sx^ + a'y, = 3a 3 - 2a\. (o) 

By solving (a) for x and y respectively we see that x x may 

become infinite, but that y x is always positive and less than a. 

Hence, to make the coefficient of y in (fi) finite for x x = oo , 

we must divide by x*, which reduces the equation of the 

asymptote to 

y = 0. 

Hence the axis of x is itself an asymptote. 

95. Points of Inflection. A point of inflection is a point 
where the tangent inter- 
sects the curve at the 
point of tangency. 

It is evident from the 
figure that in passing 
along the curve, and con- 
sidering the slope of the FlG - S1 - 
tangent at each point, the point of inflection is one at which 
this slope is a maximum or a minimum. Because we have 

slope = J-, 

the conditions that the slope shall be a maximum or minimum 
are 




§ = o 



dx 

d 3 y 
and -J-? different from zero. If the first condition is fulfilled, 

ax 

d'y . 
but if -y^ is also zero, we must proceed, as in problems of maxi- 



ASYMPTOTES AND SINGULAR POINTS. 161 

ma and minima, to find the first derivative in order which 

does not vanish. If the order of this derivative is even, there 

d u v 
is no point of inflection for -j~ = 0; if odd, there is one. 

cix 

As an example, let it be required to find the points of in- 
flection of the curve 

xy 1 = c?(a — x). 
Keducing the equation to the form 

y =£-«. 

- , dy a* 

weflnd ±=~%Iy' 

The condition that this expression shall vanish is 

±xy* = a\ 

which, compared with the equation of the curve, gives, for the 
co-ordinates of the point of inflection, 

3 , a 

EXERCISES. 

Find the points of inflection of the following curves : 

*i x a (* = ««* 

1. xy = a log — . Ans. ■{ 

a (y = Ue~\ 

(x = a(l — cos u); 
\ y =z a(nu -j- sin u). 

_ (n + 1)^ 



Ans, -{ 



„„(«.-,(_ \) + i*=S) 



162 THE DIFFERENTIAL CALCULUS. 

96. Singular Points of Curves. If we conceive an infini- 
tesimal circle to be drawn round any 
point of a curve as a centre, then, in 
general, the curve will cut the circle in 
two opposite points only, which will 
be 180° apart. 

But special points may sometimes be 
found on a curve where the infinitesimal circle will be cut in 
some other way than this: perhaps in more or less than two 
points; perhaps in points not 180° apart. These are called 
singular points. 

The principal singular points are the following: 

Double-points; at which a 
curve intersects itself. Here the 
curve cuts the infinitesimal circle 
in four points (Fig. 33). 

Cusps; where two branches of 
a curve terminate by touching 
each other (Fig. 34). Here the Fia 33 - 
infinitesimal circle is cut in two coincident points. 

Stopping Points; where a curve suddenly 
ends. Here the infinitesimal circle is cut in ^-^ 

Fig. 35. 

only a single point. ^ 

Isolated Points; from which no curve proceeds, so K^y 
that the infinitesimal circle is not cut at all. fig. 36. 

Salient Points; from which proceed two branches making 
with each other an angle which is neither zero nor 180°. 
Here the infinitesimal circle is cut in two points which are 
neither apposite nor coincident. 

There may also be multiple-points, through which the curve 
passes any number of times. A double-point is a special kind 
of multiple-point. 

A multiple-point through which the curve passes three 
times is called a triple-point. 




ASYMPTOTES AND SINGULAR POINTS. 



163 



X 


* /'' 


ZNp. 




/ 
/ 

xo ' Q/ 

L -yf. 








\ ! 
y)' 

^"' • 

— 1 

1 

1 


o 


1 i 


u 









97. Condition of Singular Points. Let (x , y ) be any 
point on a curve, and let it be required to investigate the 
question whether this point is a singular one. We first trans- 
form the equation of the 
curve to one in polar co- 
ordinates having the point 
(x , y ) as the pole. To do 
this we put, in the equation 
of the curve, 

z = *o + P cos 0; ) ^ 
y = Vo+ psinO.) 

The resulting equation 
between p and 6 will be the Fig. 37. 

equation of the curve referred to (x Q , y ) as the pole. More- 
over, if we assign to p a fixed value, the corresponding value 
of 6 derived from the equation will be the angle 6 showing 
the direction QP from Q to the point P, where the circle of 
radius p cuts the curve. The limit which 6 approaches as p 
becomes infinitesimal will determine the points of intersection 
of the infinitesimal circle with the curve. 

If, now, the given equation of the curve is 

F(x, y) = 0, 

then, by the substitution (1), the polar equation will be 

F{x Q + P cos 0,y o +p sin 6) = 0. (2) 

Now, let us develop this expression in powers of p by Mac- 
laurin's theorem. Since p enters into (2) only through x and 
y in (1), we have 



dF 
dp 

because 

Then 



<W<^<Wdy_ 
dx dp dy dp 



dx 
dp 



cos 6 and - - 
dp 



JLF , . JLF ™ 
cos 0— + sm 6-=- = F', 
dx dy 

sin d\ 



164 TEE DIFFERENTIAL CALCULUS. 

d*F dF ; Jd'Fdx , d'F dy 

-J-T = -7— — C0S ^TT "J" + -7—7- T7 

dp dp \dx dp dxdy dp, 

, . ./rZ 2 ^ 7 rfz , d a Fdy\ 

Ward?/ dp rZy dp/ 

= cos" 0— ,- + 2 sm 6 cos 6 1 - - + sm' 0-=-,- = ^ v/ . 

Noting that when p = then a; = z , we see that the de- 
velopment by Maclaurin's theorem will be 

F(x, y) = Jfo, y.) + p^cos 0— + sin 0— J 

-f- -p 2 cos 2 0-—^ + 2 sm cos 3 — = \- sm 2 0^—? 

+ etc. = 0. 

Here -=- means the value of -=— when x n is put for x, etc. 
dx Q dx ° r 

Because (x of y ) is by hypothesis a point on the curve, we 
have F(x , y ) = 0, and the only terms of the second member 
are those in p, p 2 , etc. Thus the polar equation (2) of the 
curve may be written 

>//> + F^fi + F ">P 3 + etc. = 0, ) 
or F ' + F "p + ^ "'p a + etc. = 0. ) {D) 

To find the points in which the curve cuts a circle of radius 
p, we have to determine 6 as a function of p from this equa- 
tion. When p is an infinitesimal, all the terms after the first 
will be infinitesimals. Hence, at the limit, where p becomes 
infinitesimal 8 must satisfy the equation 

dF 

dx 
which gives tan 6 = — —a. 

<¥o 
This is the known equation for the slope of the tangent at 
(x Q , yj, and gives only the evident result that in general the 



ASYMPTOTES AND SINGULAR POINTS. 165 

curve cuts the infinitesimal circle along the line tangent to 
the curve at Q. 
But, if possible, let the point (x y ) be so taken that 

d f = 0; <£ = 0. (4) 

Then we shall have F/ — 0, and the equation (3) of the 
curve will reduce to 

F Q "P + Ft'"? + etc. = 0, 
or F," + F '"p + etc. = 0. 

Again, letting p become infinitesimal, we shall have at the 
limit 

F" = cos 2 6^ + 2 sin 6 cos 0-^-- + sin 2 6~ = 0. (5) 

Dividing throughout by cos 2 6, we shall have a quadratic 
equation in tan 6, which will have two roots. Since each 
value of tan 6 gives a pair of opposite points in which the 
curve may cut the infinitesimal circle, and since (5) depends 
on (4), we conclude: 

The necessary condition of a double-point is that the three 
equations 

shall be satisfied by a single pair of values of x and y. 

If the two values of tan 6 derived from F" = are equal, 
we shall have either a cusp, or a point in which two branches 
of the curve touch each other. If the roots are imaginary, 
the singular point will be an isolated point. 

98. Examples of Double-points. A curve whose equation 
contains no terms of less than the second degree in x and y 
has a singular point at the origin. For example, if the equa- 
tion be of the form 

F(x, y) - Pr + Qxy + By* = 0, 
then this expression and its derivatives with respect to x and 
y will vanish for x = and y = 0. 



166 



THE DIFFERENTIAL CALCULUS. 



Let us now investigate the double-points of the curve 

(y s - ay - 3«V - 2ax 9 = 0. 
We have 



dF 
dx 

dF 

dy 



6(a 7 x -f- ax z ) = — 6ax(a -f re); 
4 y(y' - a a ) = ±y(y +a)(y- a). 



(1) 



(^) 



The first of these derivatives vanishes for x = or — a; 
The second of these derivatives vanishes f or y = 0, — a or -f «. 

Of these values the original equation is satisfied by the fol- 
lowing pairs: 

x = 0; 0; -a;) 

y. = — «; +«; o;5 

which are therefore the co-ordinates of singular points. 



(3) 




Fig. 38. 



Differentiating again^ we have 

tTF 



d 3 F a 



12ax; 



dxdy 



0; W = VS-±a> 



Forming the equation F" = 0, it gives 



ASYMPTOTES AND SINGULAR POINTS. 167 

(12# 2 - 4a 2 ) tan 2 6 = 6a* + 12ax. 

Substituting the pairs of co-ordinates (3), we find: 

At the point (0, — a), tan 6 — ± i VS\ 

At the point (0, + a), tan 6 = ± i V3: 

At the point (- a, 0), tan 8 = ± Vf. 

The values of tan 6 being all real and unequal, all of these 
points are double-points. The curve is shown in the figure. 

Eemakk. In the preceding theory of singular points it is 
assumed that the expression (2), § 97, can be developed in 
powers of p. If the function F is such that this development 
is impossible for certain values of x Q and y , this impossibility 
may indicate a singular point at (x , y ). 

99. Curve-tracing. We have given rough figures of va- 
rious curves in the preceding theory, and it is desirable that 
the student should know how to trace curves when their 
equations are given. The most elementary method is that of 
solving the equation for one co-ordinate, and then substitut- 
ing various assumed values of the other co-ordinate in the 
solution, thus fixing various points of the curve. But un- 
less the solution can be found by an equation of the first or 
second degree, this method will be tedious or impracticable. 
It may, however, commonly be simplified. 

1. If the equation has no constant term, we may sometimes 
find the intersections of the curve with a number of lines 
through the origin. To do this we put 

y = mx 

in the equation, and then solve for x. The resulting values 
of x as a function of m are the abscissas of the points in which 
the curve cuts the line 

y — mx = 0. 
Then, by putting 

m = ± 1, ±2, etc.; m = ± ■§•, ± J-, etc., 
we find as many points of intersection as we please. 



168 THE DIFFERENTIAL CALCULUS. 

To make this method practicable,, the equations which we 
have to solve should not be of a degree higher than the second. 

If the curve has a double-point, it may be convenient to 
take this point as the origin. 

2. If the equation is symmetrical in x and y or x and — y, 
the curve will be symmetrical with respect to one of the lines 
x — y = and x + y — 0. 

The equation may then be simplified by referring it to 
new axes making an angle of 45° with the original ones. 

The equations for transforming to such axes are 
x = (x f -j- ■(/') sin 45°; 
y — {x' — y') sin 45°. 

Application to the Folium of Descartes, If, in the equa- 
tion of this curve, 

x 3 -f y 3 = 3axy, 

we put y = mx, we shall find 

3 am 3 am* 



y = 



We also find, from the equation of the curve and the pre- 
ceding expressions for x and y in terms of m, 
dy _ z 3 — ay __ 2m — m* 



Then, for 



dx ax — 7/ 1 — 2m s ' 
3 3 



m = 1, .3 = £«; y = £- a; 

wi = 2, x = -a; y = ^a; 

3 36 54 

m= r X = 35 a; y= 35^ 

m = — 2, z = ya; # = — y«; 
etc. etc. etc. 



cfe 


= - 


- 1. 


rfy 




4 


£?£ 




5* 


dx 


= 


33 
92* 


dy 
~dx~ 


etc. 


20 
If 



Thus we have, not only the points of the curve, but the 
tangents of the angle of direction of the curve at each point, 
which will assist us in tracing it. 



THEORY OF ENVELOPES. 169 



CHAPTER XIV. 

THEORY OF ENVELOPES. 

100. The equation of a curve generally contains one or 
more constants, sometimes called parameters. For example, 
the equation of a circle, 

(x - af + (y - If = r', 

contains three parameters, a, b and r. 

As another example, we know that the equation of a 
straight line contains two independent parameters. 

Conceive now that the equation of any line, straight or 
curve, (which we shall call "the line" simply,) to be written 
in the implicit form 

cp(x } y, a) = 0, (1) 

a being a parameter. By assigning to a the several values 
a, a f , a" ', etc., we shall have an equal number of lines whose 
equations will be 

0(s> V> a ) = 0; <P(x, y, tx') =:= 0; <f>(x, y, a") = 0; etc. 

The collection of lines that can thus be formed by assign- 
ing all values to a parameter is called a family of lines. 

Any two lines of the family, e.g., those which have a and 
a' as parameters, will in general have one or more points of 
intersection, determined by solving the corresponding equa- 
tions for x and y. The co-ordinates, x and y, of the point of 
intersection will then come out as functions of a and a'. 

Suppose the two parameters to approach innnitesimally near 
each other. The point of intersection will then approach a 
certain limit, which we investigate as follows: 



170 THE DIFFERENTIAL CALCULUS. 

Let us put 

a' = a -f- Aa a 

The equations of the lines will then be 

cp(x, y, a) — and (p(x, y, a -f A a) = 0. 

If we develop the left-hand member of the second equation 
in powers of A a by Taylor's theorem, it will become 

0(z, y, a) +^a + -^ -^ + etc. = 0. 

Subtracting the first equation, dividing the remainder by 
A a, and passing to the limit, we find 

d(p{x, y, a) _ Q 

Hence the limit toward which the point (x, y) of intersec- 
tion of two lines of a family approaches as the difference of 
the parameters becomes infinitesimal is found by determining 
x and y from the equations 

«*, 9 ,a) = and '*** «> = 0. (2) 

The values of & and ?/ thus determined will, in general, be 
functions of a; that is, we shall have 

*-/,(«); y =/.(«); (3) 

which will give the values of the co-ordinates x and y of the 
limiting point of intersection for each value of a, 

Now, suppose a to vary. Then x and y in (3) will also 
vary, and will determine a curve as the locus of x and y. 

Such a curve is called the envelope of the family of 
lines, <p(x, y, a) = 0. 

In (3) the equations of the curve are in the form of (2), 
§ 76, a being the auxiliary variable. By eliminating a either 
from (2) or (3), we have an equation betAveen x and y which 
will be the equation of the curve in the usual form. 



THEORY OF ENVELOPES. Ill 

101. Theorem. The envelope and all the lines of the 
family which generate it are tangent to each other. 

Geometrically the truth of this will be seen by drawing a 
series of lines varying their position according to any con- 
tinuous law, as in the first example of the following sec- 
tion. Taking three consecutive lines and numbering them 
(1), (2) and (3), it will be seen that as (1) and (3) approach 
(2) their points of intersection with (2) approach infinitely 
near each other. Since these infinitely near points of inter- 
section also belong to the envelope, the line (2) passes through 
two infinitely near points of the envelope and is therefore a 
tangent to the envelope. 

Analytic Proof. The equation of the envelope is found by 
eliminating a from the equations (2), and we may conceive 
this elimination to be effected by finding the value of a from 
the second of these equations (2), and substituting it in the 
first equation. That is, the equation 

<P(x, y,a)=0 (4) 

represents any line of the original family when we regard a 
as a constant; and it represents the envelope when we regard 
«asa function of x and y, satisfying the equation 

d<P(x y,c,) = ^ . 

da x ' 

Let the value of a derived from this last equation be 

a = F{x, y). (6) 

Now, to find the slope of the tangent to the original line of 
the family at the point (x, y), we differentiate (4), regarding 
a as a constant. Thus we have 

tf0 dtp dy_ __ Q or dy_ = _ JW (?) 

dx dy dx dx l) y <p' 

If the original line is a straight one, this equation will give 
its slope. 

To find the slope of the tangent to the envelope at the same 



172 



THE DIFFERENTIAL CALCULUS. 



point, we differentiate this same equation, regarding a as hav- 
ing the value (6). Thus we have 

(l<P + d<pdy_ dcp/da ,dady\_. ' (s) 

dx dy dx da\dx dy dx J ' * ' 



dy 
defy 



But, because -j~ = 0, this equation will also give the value 

(7) for the slope; whence the curves have the same tangent at 
the point {x, y), and so are tangent to each other at this point. 

102. We shall now illustrate this theory by some examples. 

1. To find the envelope of a straight line which moves so that 
the area of the triangle tuhich it forms with the axes of co- 
ordinates is a constant. 




Since the area of the triangle is half the product of the 
intercepts of the axes cut off by the line, this product is also 
constant. 

Calling a and o the intercepts, the equation of the line may 
be written in the form 



0(z, y, a) = | -{- 1 - 1 = Q, 



(1) 



THEORY OF ENVELOPES. 173 

Here we have two varying parameters, a and 5, while, to 
have an envelope, the change of the parameters must depend 
on a single varying quantity. But the condition that the 
product of the intercepts shall be constant enables us to elimi- 
nate one of the parameters, say b. We have, by this condition, 

, db c 

whence -=— = „. 

da a 

Now differentiating the equation (1) with respect to a 9 re- 
garding b as a function of a, we have 

d± _ _*_ _V_db__ C V — V x _ y _ x _ _ /ox 
<fo a a 6 a rfa « 2 6 3 c " a 2 K } 

We have now to eliminate a from the equations (1) and (3), 
using (2) to eliminate b from (1). The easiest way to effect 
this elimination is as follows: 

From (3) we have 



, lex 

V = ex; a = \ / — . 

V y 



(4) 



Multiplying (1) by a, and substituting for b its value from 
(2), we have 

. a *y 

xA — a. 

c 

Substituting from (4), this equation becomes 
2x = a = a /— , 

and thus the equation of the envelope becomes 

xy = \c, 

which is that of an hyperbola referred to its asymptotes. 

This result coincides with one already found in Analytic 
Geometry, that tangents to an hyperbola cut off from the 
asymptotes intercepts whose product is a constant. 



174 



THE DIFFERENTIAL CALCULUS. 



2. To find the envelope of the line for which the sum of the 
intercepts cut off from the co-ordinate axes is a constant. 




Fig. 40. 

Let c be the constant sum of the intercepts. Then, if a be 
the one intercept, the other will be c — a. Thus the equa- 
tion of the line is 

g , V 

a 



1, 



c — a 
in which a is the varying parameter. 

Clearing of fractions, we may write the equation 
<p{x, y, a) = ex + a(y - x - c) -f a 2 = 0, 
d<p 



whence 



da 



= y — x — c-\-2a = 0, 



From the last equation we have 

a = i(x - y + c); 
this value of a being substituted in the other gives 

ex - \{x - y + c)' = 0, 
or (x - yf - 2c(x + y) -f c a = 0. 



THEORY OF ENVELOPES. 175 

This equation, being of the second degree in the co-ordi- 
nates, is a conic section. 

The terms of the second degree forming a perfect square, 
it is a parabola. 

The equation of the axis of the parabola is 

x — y = 0. 

To find the two points in which the parabola cuts the axis 
of X we put y — 0, and find the corresponding values of x. 
The resulting equation is 

x 2 — 2cx + c 2 = 0. 

This is an equation with two equal roots, x = c, showing 
that the parabola touches the axis of X at the point (c, 0). 
It is shown in the same way that the axis of Y is tangent to 
the parabola. 

It may also be shown that the directrix and axis of the 
parabola each pass through the origin, and that the parame- 
ter is VH>c, 

3. If the difference of the intercepts cut off by a line from 
the axes is constant, it may be shown by a similar process 
that the envelope is still a parabola. This is left as an exer- 
cise for the student, who should be able to demonstrate the 
following results : 

(a) When the sum of the intercepts is a positive constant, 
the parabola is in the first quadrant ; when a negative con- 
stant, the parabola is in the third quadrant. 

(/?) When the difference, a — b, of the intercepts is a posi- 
tive constant, the parabola is in the fourth quadrant; when a 
negative constant, in the second. 

(y) The co-ordinate axes touch the parabola at the ends of 
the parameter. 

In each case the parabola touches each co-ordinate axis at 
a point determined by the value of the corresponding inter- 
cept when the other intercept vanishes, and each directrix 
intersects the origin at an angle of 45° with the axis. 



176 THE DIFFERENTIAL CALCULUS. 

4. Next take the case in which the sum of one intercept 
and a certain fraction or multiple of the other is a constant. 

Let m be the fraction or multiplier. We then have 

J) -\- ma = c = a constant. 
The equation of the line then becomes 

a c — ma 

Proceeding as before, we find the equation of the envelope 

to be 

{mx — y) 2 — 2c(mx -f- y) -f c Q = 0, 

which is still the equation of a parabola. 

5. To find the envelope of a line which cuts off intercepts 
subject to the condition 

a^ + T> =1 > {a) 

m and n being constants. 

We may simplify the work by substituting for the varying 
intercepts a and b the single variable parameter a determined 
by either of the equations 

m n 

sin a — — ; cos a = — . 
a b 

The equation of the varying line will then oecome 

x 1/ 

Mx, y) = — sin a 4- — cos a — 1. (1) 

By differentiating with respect to a, we have 

d<P x i/ . 

—- = — cos a - - sm a = 0. (2) 

da on n v ' 

We may now eliminate a by simply taking the sum of the 
squares of these equations, which gives 

m* + n* ~ ±f 
the equation of an ellipse whose semi-axes are m and n. 



THEORY OF ENVELOPES. Ill 

6. To find the envelope of a circle of constant radius whose 
centre moves on a fixed circle* 

For convenience let us take the centre of the fixed circle as 
the origin, and put: 
a? b, = the co-ordinates of the centre of the moving circle; 
c = its radius; 

d = the radius of the fixed circle. 
The equation of the moving circle now becomes 

{x - a)' + {y- b)\ - e = 0. (1) 

By differentiation with respect to a, 

z-«+(2/-J)J = 0. 

The condition that (a, o) lies on the fixed circle gives 

« a + b 7 = d\ (2) 

, db a 

whence -—=—-. 

da b 

Then, by substituting this value, 

ay — bx = 0. (3) 

We have now to eliminate a and b from (1), (2) and (3). 
Firstly, from (1) and (2), we find 

as" + y % - 2ax - %by = c x - d\ (V) 

From (2) and (3) we find the following expressions for a 
and b: 

a = ; o = - 

Vx* + if Vx* + f 

By substitution in (1'), and putting for brevity 

r' = x* + f, 

we find r 5 ± 2rd + d* = c\ 

Hence r 2 = x* + y* = (c ± d)\ 

the equations of two circles around the origin as a centre, 
with radii c + d and c — d. 



178 THE DIFFERENTIAL CALCULUS. 

7. Find the envelope of a family of ellipses referred to their 
centre and axes, the product of whose semi-axes is equal to 
a certain constant, c 2 . 

Ans. The equilateral hyperbola xy = \c*. 

8. To find the envelope of a family of straight lines, such 
that the product of their distances from two fixed points is a 
constant. 

Let (a, 0) and (— a, 0) be taken as the two fixed points, 
and let c 2 be the constant. Also, let 

x cos a -j- y sin a — p = (1) 

be the equation of any one of the lines in the normal form, 
p and a being the varying parameters. 

The distances of the line from the points {a, 0) and ( — a, 
0) are respectively 

— p -\- a cos a and — p — a cos a. 

Hence we have the condition 

p* — a 2 cos 2 a = c*. (2) 

Differentiating (1), regarding p as a function of a, we have 

i dp 
— x sin a + y cos a f- = 0. 

From (2) we obtain 

dp a? sin a cos <z 

<ia ^? 

We thus have the three equations 

x cos a -j- y sin a = p, (a) 

a? sin # cos a /TX 

x sin <* — y cos a = , (J) 

p 2 = c 2 + # s cosS a 

= c 2 -f- a 2 — a 8 sin 2 a, (c) 

from which to eliminate p and or. 



THEORY OF ENVELOPES. 179 

To effect the elimination of a and p we find the values of 
x and y from (a) and (b) by taking 

(a) X cos a -f (#) x sin a and («) x sin a — (£) X cos «. 
We thus find, by the aid of (c), 

px = p 1 cos a -\- a 1 sin 9 « cos or; 
, a , sx cos a 

, sin ^ 
y = c 



Hence 



P 
x cos a 



c* + a* p ' 
y _ sin a 
c* ~ J3 

If we multiply the first of these equations by x and the 
second by y and add, then we have 

r* y* _ x cos a -f y sin a _ 

Tl — ~ — ■*■• 



c' + a a ' e a jt? 

Hence the equation of the envelope is 

c * _|_ a % ^ c a 

This represents an ellipse whose foci are the two fixed 
points. 

This interpretation, however, presupposes that the product 
c a of the distances of the line from the two points is positive; 
that is, that the points are on the same side of the enveloping 
line. If the product is negative, the equation of the envelope 
will be 

* _ t - i 

a » - e &~ 9 

which is the equation of an hyperbola. 

These results give the theorem of Analytic Geometry that 
the product of the distances of a tangent from the foci of a 
conic is constant. 



180 THE DIFFERENTIAL CALCULUS. 



CHAPTER XV. 

OF CURVATURE; EVOLUTES AND INVOLUTES. 

103. Position; Direction; Curvature. The position of 
any point P on a curve is fixed by the values of the co-ordi- 
nates, x and y, of P. This is shown in Analytic Geometry. 

If we have given, not only x and y, but the value of -—- for 

the point P, then such value of the derivative indicates the 
direction of the curve at the point P, this direction being the 
same as that of the tangent at P. 

The curve may also have a greater or less degree of curva- 
ture at P. The curvature is indicated by a change in the di- 

dti 
rection of the tangent, that is, in the value of —-, when we 

ax 

pass to an adjacent point P\ But such change in the value 

dv 
of ~- when we vary x is expressed by the value of the second 
ax 

d*y 
derivative -~. If this quantity is positive, the angle which 

the tangent makes with the axis of JTis increasing with x at 

the point P, and the curve, viewed from below, is convex. 

d*y 
If -r-4 is negative, the tangent is diminishing, and the 

ax 

curve, seen from below, is concave. 

To sum up: If we take a value of the abscissa x, then the 
corresponding value of 

y gives the position of a point P of the curve; 

~- gives the direction of the curve at P; 

d*y 

-^4 depends upon the curvature of the curve at P. 

CLX 



CURVATURE; E VOLUTES AND INVOLUTES. 181 

104. Contacts of Different Orders. Let two different 
curves be given by their respective equations: 

y =f( x ) and y = 00*0- 

If for a certain value of x, which value call x , the two 

values of y are equal, the two curves have the corresponding 

point in common; that is, they meet at the point (x , y). 

clii 
If the values of -— are also equal at this point, it shows 

that the curves have the same direction at the point of meet- 
ing. They are then said to touch each other. 

d*y 
If the values of -~ are also equal at this point, the two 

(JLX 

curves have also the same curvature at this point. 

To show the result of these several equalities, let us give 

the abscissa x (which we still take the same for both curves), 

an increment h, and develop the two values of y in powers of 

civ 
h by Taylor's theorem. To distinguish the values of y, ~ 9 

etc., which belong to the two curves, we assign to one the 
suffix 0, and to the other the suffix 1. Then, for the one 
curve, 



and, for the other, 

»--'■+ (I) *+(£),&+••■ + ©.5+"" 

The difference between the values of y f and y is the inter- 
cept, between the two curves, of the ordinate at the point 
whose abscissa is x -f- h. Its expression is 
Uly' 



Vx -y t 



dx) l \dxJJ \SdxV, \dxVj 1*2 



etc. 



Now, consider the case in which the curves meet at the 
point P, whose abscissa is x . Then 

yi-y* = o, 



182 THE DIFFERENTIAL CALCULUS. 

and the intercept of the ordinate will be 

M)r @r) ] h + terms in w > ei0 -' 

which, when h becomes infinitesimal, is an infinitesimal of 
the first order. 
If we also have 



ldy\ = !dy\ 
\dx I , \dx I , 



the ordinates will differ only by a quantity containing h* as a 
factor, and so of the second order. Hence: 

WJien two curves are tangent to each other, they are sepa- 
rated only hy quantities of at least the second order at an in- 
finitesimal distance from the point of tangency. 

In the same way it is shown that if the second differential 
coefficient also vanishes, the separation will be of the third 
order, and so on. 

Def. When two curves are tangent to each other, if the 
first n differential coefficients for the two curves are equal at 
the point of tangency, the curves are said to have contact 
of the nth order. 

Hence a case of simple tangency is a contact of the first 
order. If the second derivatives are also equal, the contact 
is of the second order, and so forth. 

105. Theorem. In contacts of an even order the two 
curves intersect at the point of contact ; in those of an odd 
order they do not. 

For, in contact of the nth order, the first term of y' — y 
(§ 104) which does not vanish contains h n+1 as a factor. 

If n is odd, n-\-\ is even, and y' — y has the same alge- 
braic sign whether we take h positively or negatively. Hence 
the curves do not intersect. 

If n is even, n -f- 1 is odd, and the values of y' — y have 
opposite signs on the two sides of the point of contact, thus 
showing that the curves intersect. 



CURVATURE; EVOLUTES AND INVOLUTES. 183 



106. Radius of Curvature, The curvature at any point 
is measured thus: We pass from the point P to a point P' in- 
finitesimally near it. The 
curvature is then measured 
by the ratio of the change 
in the direction of the tan- 
gent (or normal) to the 
distance PP'. Let us put 
a = the angle which 
the tangent at P makes 
with the axis of X. 

a -J- da = the same angle for the tangent at P' 
ds = the infinitesimal distance PP'. 

Then, by definition, 




Fig. 41. 



Curvature = 



da 
ds* 



Now, because 
we have, by differentiation, 



tan a = -^- 9 
dx 



sec Q a da 



dx' 



dx. 



Also, 



sec' a = 1 4- tan 2 a = 1 -j- 



df 
dx' 



and ds = y (l + -j-Adx. 

From these equations we readily derive 

~ , da dx* 

Curvature = — - = 
ds 



Now, draw normals to the curve at the points P and P', 
and let C be their point of intersection. Because they are 
perpendicular to the tangents, the angle PCP f between them 
will be da, and if we put 



184 THE DIFFERENTIAL CALCULUS. 

we shall have 



PP' = ds = pda. 



Hence p = — 



(>+»■ 



da curvature d 2 y 

dx~* 

The length p is called the radius of curvature at the 
point P, and C is called the centre of curvature. 

Cokollaey. The centre of curvature for any point of a 
curve is the intersection of 
consecutive normals cut- 
ting the curve infinitely 
near that point* v\ 




107. The Osculating 
Circle. If, on the normal 
PC to any curve at the 
point P, we take any point Fig ^ 

as the centre of a circle 

through P, that circle will be tangent to the curve at P; 
that is, it will, in general, have contact of the first order 
at P. But there is one such circle which has contact of a 
higher order, namely, that whose centre is at the centre of 
curvature. Since this circle will have the same curvature 
at P as the curve itself has, it will have contact of at least 
the second order at P. 

This proposition is rigorously demonstrated by finding that 
circle which shall have contact of the second order with the 
curve at the point P. 

Let us put 

x, y, the co-ordinates of P\ 



p = ~ for the curve at the point P; 
a = -~ for the curve at the point P, 






QXm¥£Wm$ EVOLUTES AND INVOLUTES. 185 

I'hese last two quantities are found by differentiating the 
equation of the curve. 

NoW> -j- and yf must have these same values at the point 

\x, y) in the case of the circle having contact of the second 
order (§ 104). 

Let the equation of this circle be 

(x - a)' + (y- b)' = r\ (a) 

By differentiation, we have 

(x — a)dx -f (y — b)dy = 0, 

whence —- = ^ = p. (b) 

ax b — y r v ' 

Differentiating again, 

ay__l_ (x-a ) dy_ _ (y-by+(x-ay 

dx* b-y^ (b-y) 2 dx ' (y - b) 3 

From (b) combined with (a) we find 

, , (x - a)' r' 



1+f 



1 (y - *)• (y - i)" 

Dividing this by (c) gives 

? ' 

the equivalent of the expression already found for the radius 
of curvature. 

Hence if we determine a circle by the condition that it 
shall have contact of the second order with the curve at the 
point P, its radius will be equal to the radius of curvature. 
TMs circle is called the osculating circle for the point P. 
Each point of a curve has its osculating circle, determined 
by the position, direction and curvature at that point. 



186 THE DIFFERENTIAL CALCULUS. 

Cor. The osculating circle will, in general, intersect the 
curve at the point of contact, for it has contact of the second 
order. 

This may also be seen by reflecting that the curvature of a 
curve is, in general, a continuously varying quantity as we 
pass along the curve, and that, at the point of contact, it is 
equal to the curvature of the circle. Hence, on one side of 
the point of contact, the curvature of the curve is less than 
that of the circle, and so the curve passes without the circle; 
and on the other side the curvature of the curve is greater, 
and thus the curve passes within the circle. 

If, however, the curvature should be a maximum or a 
minimum at the point of contact, it will either increase on 
both sides of this point or diminish on both sides, whence 
the circle will not intersect the curve. 

108. Radius of Curvature tohen the Abscissa is not taken 
as the Independent Variable. Suppose that, instead of x, 
some other variable, u, is regarded as the independent vari- 
able. We then have 

y=f(u); x=f 2 (u). 

Now, it has been shown that, in this case, we have (§56) 



d*y dx d 7 x dy 
d*y __ du 2 du du 2 du 

\du I 
Also, we have 

(dyV fdxV td^\ 
\duJ _\duj__\du) 
(dxV ~ fdx\ 

\du I \du J 



dx 2 ~~ W 



i . l dl )\ -. , W" V*«/ ™*/ ,a^ 

1 + \dx~J = 1 h ^^ = ^~^~ (3) 



These expressions being substituted in the expression for 
the radius of curvature, it becomes 



CURVATURE j EV0LUTE8 AND INVOLUTES. 187 

2 \ § 



{(£)•+ ffi)'} 



109. Radius of Curvature of a Curve referred to Polar 
Co-ordinates. Let the equation of the curve be given in the 
form 

r = <p{6). 

The preceding expression (4) may be employed in this case 
by taking the angle 6 as the independent variable. By differ- 
entiating the expressions 

x = r cos 8, 
y = r sin 6, 

regarding r as a function of 6, we find, when we put, for 
brevity, 

, _ dr „ __ d/r 

r = dd' r ~w 

—j. = — r sin 6 -f r' cos 0; 
^ = ( r « _ r ) cos 61 - % r ' sin 0; 

-^ = r cos 6* -f r' sin d; 
<Fy _ ( r ./ _ r ) sin # + 2r ' C os ft 

By substituting these derivatives with respect to & for those 
with respect to u in (4) and performing easy reductions, we 
find 



d0 9 



^' + 2 U) 



which is the required expression for the radius of curvature. 



188 THE DIFFERENTIAL CALCULUS 

EXAMPLES AND EXERCISES. 

1. The Parabola. To find the radius of curvature of a 
curve at any point, we have to form the value of p from the 
equation of the curve. The equation of the parabola is 

y' = 2p X , 

whence we find 

cly _ p m 
dx ~ y ' 
cVy_ _f 
dx' y 3 ' 

Then, by substituting in the expression for p, we find 

„ _ (£ + p')* 
f>- f , 

the negative sign being omitted, because we have no occasion 
to apply any sign to p. 

At the vertex y = 0, whence 

p=p. 

Hence, at the vertex, the radius of curvature is equal to 
the semi-parameter, and the centre of curvature is therefore 
twice as far from the vertex as the focus is. 

2. Show that the radius of curvature at any point {x, y) of 
an ellipse is 

_ (aV + av) f 

P ~ a*V 

and show that at the extremities of the axes it is a third pro- 
portional to the semi-axes. 

3. Show that the algebraic expression for p is the same in 
the case of the hyperbola as in that of the ellipse. 

4. What must be the eccentricity of an ellipse that the cen- 
tre of curvature for a point at one end of the minor axis may 
lie on the other end of that axis? Ans. e — \/%. 



CURVATURE; EV0LUTE8 AND INVOLUTES. 189 

5. Show that in the case supposed in the last problem the 
radius of curvature at an end of the major axis will be one 
fourth that axis. 

6. The Cycloid. By differentiating the equations (1), § 80, 
of the cycloid, we find 

dx 



-Y- = a — a cos u = y\ 
du y ' 

d*x dy 

du du 

d 2 y 

■—= = a cos u. 

du 



(2) 



Then, by substituting in (4) and reducing, we find, for the 
radius of curvature, 

p = 2*# Vl — cos u = 4« sin \u . 

We see that at the cusp, 0, of the cycloid, where u = 0, 
the radius of curvature also becomes zero. 

7. The Archimedean Spiral. Show from (5) that the ra- 
dius of curvature of this spiral (r = ad) is 

' 2 + 0" * 

8. The Logarithmic Spired. The equation of the loga- 
rithmic spiral being 

19 

r = ae , 
show that the radius of curvature is 



p = r VT+l\ 

Hence show that the line drawn from the centre of curva- 
ture of any point P of the spiral to the pole is perpendicular 
to the radius vector of the point P. 

9. Show that the radius of curvature of the lemniscate in 
terms of polar co-ordinates is 

a a 2 



190 



THE DIFFERENTIAL CALCULUS. 



110. Evolutes and Involutes. For every point of a curve 
there is a centre of curvature,, found by the preceding for- 
mulae. The locus of all such 
centres is called the evolute 
of the curve. 

To find the evolute of a 
curve, let (x s y t ) be the co-ordi- 
nates of any point P of the 
curve ; PC, the radius of cur- 
vature for this point; and a, 
the angle which the tangent 
at P makes with the axis of X. 
Then, for the co-ordinates of 
(7, we have 

x = Xj — 




Fig. 43. 



p sm a; 

y = y, + p<x>8 a. 

Substituting for p its value (§ 106), and for sin a and cos a 
their values from the equation 

dx? 



tan a 



we find 



1 + 



x = x, 



dx. 



y = y,+ 



*1a 

dx/ 

! + */ 

3 dx; 
dx; 



dyj. 
dx; 



(i) 



If in the second members of these equations we substitute 
the values of the derivatives obtained from the equation of 
the curve, we shall have two equations between the four vari- 
ables x, y, x t and y r By eliminating x 4 and y from these 
equations and that of the given curve, we shall have a single 
equation between x and y, which will be that of the evolute. 



CURVATURE; EV0LUTE8 AND INVOLUTES. 191 

111. Case of an Auxiliary Variable. If the equation of 
the curve is expressed by an auxiliary variable, we have to make 

in (1) the same substitution of the values of -—, -=-— * , etc., 

v ' du du 

as in § 108. Thus we find, instead of (1), 



[dx t 

dy t \du 



+ 



du 



du d*y 4 dx 4 d?x 4 dy/ 
du* du du* du 



dx t 



\du 1 ^ \ 



(fyj 

du 



du d*y 4 dx t d*x t dy 4 ' 
du* du du? du 



(3) 



which are the equations of the evolute in the same form. 



EXAMPLES OF EVOLUTES. 

112, The Evolute of the Parabola. If we substitute in 
(1) for the derivatives of y t with respect to x t the values 
already found for the parabola, these equations (1) become 

* = */+*+£ =*+lf> 

y = - K 

"We now have to eliminate y t from these two equations, % t 
having already been eliminated by the equation of the curve. 
They give 

y/ = ip(x-p); y/ = -fy- 

Equating the cube of the first equation to the square of the 
second, we find, for the equation of the evolute of the parabola, 



y = * 



(x - pY 



27 p 



192 



THE DIFFERENTIAL CALCULUS. 



113. Evolute of the Ellipse. From the equation of the 
ellipse, we find 

dy s _ b*x^ d*y s _ b* 

dx t ~ a*y/ dx/ ~ d 2 y*' 
By substituting in (1) and reducing, we find 

x - x Mid i ^V^ ^ 3 -^/-M / 

Remarking that the equation of the ellipse gives 
a 4 b* - a 4 y/ = a*Vx\ 
and putting e 2 = a? — b*, 

the preceding equation becomes 

* = ^& (•) 



In the same way we get 

9 = 9,-- jt-[i + ?£) = — fc 



<») 



In this case the easiest way to effect the elimination of x t 
and y t is to obtain the values of these quantities from (a) 
and (b), and then substitute them in the equation of the 
ellipse. From (a) and (b), we find 



-®t<*~-®}- 



which values are to be 
substituted in the equa- 
tion 



a* "*" b' 



1. 




We thus find, for the 
equation of the evolute of 
the ellipse, 

a¥ + b l y l = c ! . 

The figure shows the 
form of the curve. The following properties should be de- 
duced by the student. 



CURVATURE; EV0LUTE8 AND INVOLUTES. 193 



y + (iy = *** - c ° s •* 

- a 2 (l — cos w); 



dii* die 



(a) The evolute lies wholly within the ellipse, or cuts it (as 
in the figure), according as e 2 < i or e* > -J. 

(b) The ratio Ci) : AB (which lines we may call axes of 
the evolute) is the inverse of the ratio of the corresponding 
axes of the ellipse. 

114. Evolute of the Cycloid. Here we have to apply the 
formulae (2) for the case of a separate independent variable. 
Substituting in (2) the values of the derivatives already given 
for the cycloid, we shall find 

fdx_ 

\du 
d*y dx 
• du 2 du 

x = x t -j- 2a sin u = a(u -j- sin u); 

y — y t — 2a(l — cos u) = — a(l — cos u). 

These last two equations are those of the evolute. 

Let us investigate its form. For u = we have x = 
and y = 0, whence the 
origin is a point of the 
curve. 

For u = n we have 

x = an; 
y = -2a; 

giving a point C, below 
the middle of the base of 
the cycloid, at the dis- 
tance 2a. Let us take this point as a new origin, and call 
the co-ordinates referred to it x r and y f . We then have 

x' = x — an = a{6 — n -j- sin 6); 
y' =zy+ 2a = a(l + cos 6). 
If we now put 





Y 


Y 




<x 




\ 


/B X 


" 


-•^ 


-•"*" 


""s 












•>. 


s 






V 


s 






















\ 


/ 


, 






/ 


___s; 




C 








Fig 


45. 





0' = 



these equations become 



194 



THE DIFFERENTIAL CALCULUS. 
x' = a(6' - sin 6'); 



y' = a(l - cos 0'); 
which are the equations of another cycloid, equal to the 
original one, and similarly situated. The cycloid therefore 
posesses the remarkable property of being identical in form 
with its own evolute. 

115. Fundamental Properties of the Evolute. 

Theoeem I. The involute of a curve is the envelope of its 
normals. 

As we move along a curve,, the normal will be a straight 
line moving according to a certain law depending upon the 
form of the curve. This line will, in general, have an en- 
velope, which envelope will be, by definition, the locus of the 
point of intersection of consecutive normals. But this point 
has been shown to be the centre of curvature, whose locus is, 
by definition, the evolute. 
Hence follows the theorem. 

Cokollaky. The nor- 
mals to a curve are tan- 
gents to its evolute. For 
this has been shown to be 
true of a moving line and 
its envelope. 

Theoeem II. If the os- 
culating circle move around 
the curve, the motion of its 
centre is along the line join- 
ing that centre to the point 
of contact. 

This theorem will be 
made evident by a study 
of the figure. If the line 
P 2 (7 2 be one of the nor- 
mals from the point of contact P 2 to the centre, then, since 




Fio. 46. 



CURVATURE ; EVOLUTES AND INVOLUTES. 195 

tills normal is tangent to the locus of the centre, it will be the 
line along which the centre is moving at the instant. 

Theokem III. The arc of the evolute contained between 
any two points is equal to the difference of the radii of the 
osculating circles whose centres are at these points. 

For, if we suppose the points C l9 C if etc., to approach in- 
finitesimally near each other, then, since the infinitesimal 
arcs G X C^ O^C 3 , etc., are coincident with those successive 
radii of the osculating circle which are normal to the curve, 
these radii are continually diminished by these same infini- 
tesimal amounts. 

The analytic proof of Theorems II. and III. is as follows: 
Let the equation of the osculating circle be 
(x - a y + {y- iy = p>, 
where a and b are the co-ordinates of the centre of curvature, 
and therefore of a point of the evolute. 

The complete differential of this equation gives 

(x — a) (dx — da) -f- (y — b) (dy — do) — pdp. (a) 
If, in this equation, we suppose x and y to be the co-ordi- 
nates of the point of contact of the circle with the curve, then 
dx and dy will have the same value at this point whether we 
conceive them to belong to the circle, supposed for the mo- 
ment to be fixed, or to the curve. But in the fixed circle we 

have 

(x — a)dx + (y - b)cly = 0. (b) 

Subtracting this equation from (a) and dividing by p, we find 

da + db — — dp, (c) 

p p 

which is a relation between the differential of the co-ordi- 
nates of the centre and the differential of the radius. Now, 
if we put /3 for the angle which the normal radius makes 
with the axis of X, we have 

x — a y — a . „ t , x 

— = cos p; = sm p. id ) 

p p 



196 THE DIFFERENTIAL CALCULUS. 

But this same normal radius is a tangent to the evolute. 
If we call a the arc of the evolute, we find by a simple con- 
struction da = cos /3da; db = sin fida. 

Multiplying these equations by cos /? and sin P, respectively, 
and adding, we find 

da = cos /3da + sin fidb. 

Comparing (c) and (d), we find 

da = — dp, 
or d(a -j- p) = 0. 

Now, a quantity whose differential is zero is a constant. 
Hence we always have 

a -\- p = constant, 
or a = constant — p. 

If we represent by a 1 and <x 2 the arcs from any arbitrary 
point of the involute to the two chosen points, and by p x and 
p 2 the values of p for these points, we have 

<r x = const. — p x ; 
cr, = const. — p % . 

. • . a n — a 1 = p 1 — p 2f 

or the intercepted arc equal to the difference of the radii, as 
was to be proved. 

It must be remarked, however, that whenever we pass a 
cusp on the evolute, we must regard the arc as negative on 
one side and positive on the other. In the case of the ellipse, 
for example, those radii will be equal which terminate at 
equal distances on the two sides of any cusp, as A, B, C or 
D, and the intercepted arc must then be taken as zero. 

116. Involutes. The involute of a curve C is that 
curve which has C as its evolute. 

The fundamental property of the involute is this: The 
involute may be formed from the evolute by rolling a tangent 



CUBVATUBUE ; EV0LUTE8 AND INVOLUTES. 197 

line upon the latter. A point P on the rolling tangent will 
then describe the involute. 

This will be seen by reference to Fig. 46. The rolling line, 
being tangent to the evolute, coincides with the radius P 1 C x , 
and as it rolls along the evolute into successive positions, 
P 3 (7 2 , P 3 C 3 , etc., the motion of the point P is continually 
normal to its direction. 

It will also be seen that the radius of curvature of the in- 
volute at each point is equal to the distance PC from P to 
the point of contact with the evolute. 

The conception may be made clearer by conceiving the 
rolling line to be represented by a string which is wrapped 
around the evolute. The involute is then formed by the mo- 
tion of a point on the string. 

The general method of determining the involutes of given 
curves involves the integral calculus. 



PART II. 
THE INTEGRAL CALCULUS. 



PART II. 
THE INTEGRAL CALCULUS. 



CHAPTER I. 

THE ELEMENTARY FORMS OF INTEGRATION. 

117. Definition of Integration. Whenever we have given 
a function of a variable x, say 

u = F(x), 
we may, by differentiation, obtain another function of x, 
du _,, . 

7S = F{x) > 

which we call the derived function. 

In the integral calculus we consider the reverse process. 
We have given a derived function 

t"(x) 9 

and the problem is: What function or functions, when differ- 
entiated, ivill have F'{x) as their derivative? 

Every such function is called an integral of F'(x). 

The process of finding the integral is called integration. 

The operation of integration is indicated by the sign / , 

called "integral of," written before the product of the given 
function by the differential of the variable. Thus the ex- 
pression 

*F'{x)dx 



/• 



means: that function whose differential with respect to x is 
F f (x)dx. 



202 THE INTEGRAL CALCULUS. 



Calling u the required f unction, then if we have 



we must also have 

As examples: 

Because d(x*) = 2xdx, 

we have / 2xdx = x*. 

Because d(ax* -\- bx -f- c) = {2ax -j- b)dx, 

we have / (2ax -f- b)dx = ax* + bx -J- c. 

And, in general, if, by differentiation, we have 
dF{x) = F'{x)dx, 

we shall have I F'{x)dx — F{x). 

118. Arbitrary Constant of Integration. The following 
principle is a fundamental one of the integral calculus: 

If F(x) is the integral of any derived function of the va- 
riable x y then every function of the form 

F{x) + h, 

h being any quantity whatever independent of x, will also be 
an integral. 

This follows immediately from the fact that h will dis- 
appear in differentiation, so that the two functions 

F(x) and F(x) + h 

have the same derivative (cf. § 24). 

The same principle may be seen from another point of 
view : Since the problem of differentiation is to find a func- 
tion which, being differentiated, will give a certain result, 
and since any quantity independent of the variable which 
may be added to the original function will have disappeared 
by differentiation, it follows that we must, to have the most 



TEE ELEMENTARY FORMS OF INTEGRATION. 203 

general expression for the integral, add this possible but un- 
known quantity to the integral. 

The quantity thus added is called an arbitrary constant. 
But it must be well understood that the word constant merely 
means independent of the variable with reference to which 
the integration is 'performed. 

It follows from all this that the integral can never be com- 
pletely found from the differential equation alone, but that 
some other datum is needed to determine the arbitrary con- 
stant and thus to complete the solution. 

Such a datum is the value of the integral for some one 
value of the variable. Let F(x) -f h be the integral, and let 
it be given that 

when x = a, then the integral = K. 

We must have, by this datum, 

F(a) + h = K, 
which gives h = K — F(a), 

and thus determines h. 

Eemark. Any symbol may be taken to represent the ar- 
bitrary constant. The letters c and h are those most gener- 
ally used. We may affix to it either the positive or the nega- 
tive sign, and may represent it by any function of arbitrary 
but constant quantities which we find it convenient to intro- 
duce It is often advantageous to write it as a quantity of the 
same kind as the variable which is integrated. 

119. Integration of Entire Functions. 
Theorem I. The integral of any power of a variable is 
the power higher by unity, divided by the increased exponent. 
In symbolic language, we have 



/ 



,£71+1 

x n dx = ■ - -f ho 
n-\-l 



For, by differentiating the expression — - — - + h, we have 



204 THE INTEGRAL CALCULUS. 

Theorem II. Any constant factor of the given differen- 
tial may be written before the sign of integration. 
In symbolic language, 



JaF'{x)dx = aJ*F\x)dx. 



This is the converse of the Theorem of § 23. By that 
theorem we have 

d(aF(x)) = adF(x), 

from which the above converse theorem at once follows. 
In the special case a = — 1 we have 

J- F\x)dx = j*F:{x)d(- x) = - fF'{x)dx. 

Hence the corollary: If the integral is preceded by the nega- 
tive sign we may place thai sign before either the derived 
function or the differential. 

Theorem III. If the derived function is a sum of several 
terms, the integral is the sum of the separate integrals of the 
terms. 

In symbolic language, 

Ax + Y+Z+ . . .)dx = f Xdx+ f Ydx+ f Zdx+ . . 

This, again, is the converse of Theorem II of § 22. 

The foregoing theorems will enable us to find the integral 
of any entire function of a variable. To take the function in 
its most general form, let it be required to find the integral 

u = I (ax m -j- bx n + cx p + . . ,)dx. 

By Theorem III., 

y, = / ax m dx -f / bx n dx -\- j cx p dx -f- • , 



THE ELEMENTARY FORMS OF INTEGRATION. 205 
By Theorem II., 



fax m dx 

etc. 
and by Theorem I., 

J x m dx = 

etc. 
By successive substitution 
ax m+1 , bx n+1 , cx p 


= a 1 x m dx; 
etc.; 

■ \ h • 


etc. 
we then have 

— - + ah. + #A, + ch z -|- 



where 7^ 1 , 7i a , h z , etc., are the arbitrary constants added to the 
separate integrals. 

Since the sum of the products of any number of constants 
by constant factors is itself a constant, we may represent the 
sum ah x \ bli^ -f ch % by the single symbol h. Thus we have 



/ 



(ax m + bx n 4- cx p + . . ,)dx 

~ m + I + M4-l~ r jt?4-l + "' + 

EXERCISES. 

Form the integrals of the following expressions, multiplied 



by dx: 














I. x\ 


2. 


z 9 . 


3. 


X~\ 


4. 


z- 3 . 


5. az". 


6. 


^ 3 . 


7. 


ax~ % . 


8. 


fo;- 3 . 


9. az 4- 5. 


10. 


«£ a — c. 


II. 


ax* 4- ex. 


12. 


rt£ s — ex 9 


13. a*. 


14. 


<A 


*5- 


x-K 


16. 


ax-%. 


17. axi—bx-i. 


18. 


X* 


19. 


a b 
x* ~ x 5 ' 


20. 


X 



120. The Logarithmic Function. An exceptional case 
of Theorem I. occurs when n = — 1, because then n 4- 1 
= 0, and the function becomes infinite in form. But since 

^•log x = — - = x~ l dx, 
x 



206 THE INTEGRAL CALCULUS. 

it follows that we have for this special case 

/ x~ x dx — I — = log x -f h. (a) 

Let c be the number of which h is the logarithm. We then 
have 

log x-\-h = log x -j- log c — log ex. 

We may equally suppose 

h = — log c = log — . 
c 

X 

Then log x -j- h = log -. 

Hence we may write either 

Pdx , I 

j-=logcx, 

/dx , a; 
— =log-; 
a; ° c 



(i) 



c being an arbitrary constant. 

We thus have the principle: The arbitrary constant added 
to a logarithm may be introduced by multiplying or dividing 
by an arbitrary constant the number whose logarithm is ex- 
pressed. 

121. We may derive the integral (a) directly from Theo- 
rem I., thus: In the general form 

x n dx = ^-— + h 
n -\- 1 

let us determine the constant h by the condition that the in- 
tegral shall vanish when x has some determinate value a. 
This gives 

a n+1 a n+1 

-f h = 0; .*. h 



n -J- 1 ' ' n + 1 

Thus the integral will become 

n+"l ' 



/ x n dx 



THE ELEMENTARY FORMS OF INTEGRATION. 207 

in which a takes the place of the arbitrary constant. This 
expression becomes indeterminate for n = — 1. But in this 
case its limit is found by § 71, Ex. 5, to be log x — log a. 
Thus we have 



/ 



x n dx = log x — log a = log — , 



as before, log a being now the arbitrary constant. 

122. Exponential Functions. Since we have 
d(a x ) = log a . a x dx, 
it follows that we have 



/ log a . ofdx — a x -\- h, 



or, applying Th. II., § 119, to the first member and then di- 
viding by log a, 

a x + h 



/ 



a x dx 



log a' 
which we may write in the form 



/* 



ar 

dx = = f- 7i, 

log a 



because - } ■ is itself a constant which we may represent by h. 

123. The Elementary Forms of Integration. There is 
no general method for finding the integral of a given differen- 
tial. What we have to do, when possible, is to reduce the 
differential to some form in which we can recognize it as the 
differential of a known function. For this purpose the fol- 
lowing elementary forms, derived by differentiation, should 
be well memorized by the student. We first write the prin- 
cipal known differentials, and to the left give the integral, 
found by reversing the process. For perspicuity we repeat 
the forms already found, and we omit the constants of in- 
tegration. 



208 THE INTEGRAL CALCULUS. 

•.•%»+>) = (n + l)y"dy, .-.fy'dy =|^. (1) 

• . • ^-sin y = cos ydy, . • .y cos ydy = sin y. (3) 

• . * d-coa y — — sin #%, . • . /sin y% = — cos y. (4) 

• . • d' tan 2/ = sec 2 ydy, . • . f. — IL = tan y. (5) 

«/ cos 2 y ./ W 



•d'coty =— -f-j!— 9 .-. I - 

sm 2 v t/ si 



sm' # f/ sm 2 y 



= - cot y. (6) 



* cosy J ' J cosy J ' v> 

... ,. tan <-», = _|L_„ .'./j^ =tan<-> y .(10) 

'.'d-a y =a v logady, r.Ja v dy = -^— . (11) 



^•sinh (_1) z/= 



' ' v v^i = sin ll(-1)y = log ( ^ + ^^-f 11 )- ( 12 ) 

'." f?*cosli (-1) ?/= — — - — , 

• ' -/^= cos h< ~'^ = Jog (y + ty^i). (i3) 

•.•<Z-tanh ( -'V=--^-„ 

i - y 2 



. • . A-^i = tan h<- »>y = I log Lt* 
«/ 1 - y' * 2 s 1 - y" 



(14) 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 209 



CHAPTER II. 

INTEGRALS IMMEDIATELY REDUCIBLE TO THE 
ELEMENTARY FORMS. 

124, Integrals Reducible to the Form / y n dy. The fol- 
lowing are examples of how, by suitable transformations, we 
may reduce integrals to the form (1). Let it be required to find 

i(a + x) n dx. 

We might develop (a + x) n by the binomial thorem, and 
then integrate each term separately by applying Theorem III., 
§ 119. But the following is a simpler way. Since we have 
dx = d(a + x), we may write the integral thus: 

C(a + x) n d(a + x). 

It is now in the form (1), y being replaced by a -f- %. 
Hence 

J {a + xfdx = (a +^° + 1 +h. (1) 



In the same way, 

To take another step, let us have to find 



/ (a — x) n dx = — J (a— x) n d{a — x) = h — ~Z\ 

p, let us have to 

I {a + bx) n dx. 
We have 

dk = yd(bx) — jd(a -|- Jo:). 

Hence, by applying Th. II., 



210 THE INTEGRAL CALCULUS. 

"We might also introduce a new symbol, y = a -f- hx, and 
then we should have to integrate y n dy with the result in § 123. 
Substituting for y its value in terms of x, we should then have 
the result (2).* 

These transformations apply equally whether n 9 a and b 
are entire or fractional, positive or negative. 

EXERCISES. 

Find: i. / (a -f x)*dx. 2. JZ(a — xfdx. 

3. / (a — 2x) 4 dx. 4. / (a -f- a)~ 2 67a. $. / (a — x)~ 2 dx. 
I {a-\-mx)~ z dx. 7. / (a — mx)*dx. 8. / (a — mx)~*dx. 

r dx r dx r dx 

J (a+ a) 2 * IO V (a - a;) 2 ' ri V (a - 4a) 1 ' 



/ (a + a) p £7a. 13. / (« + nx) p dx. 14. / (a -J- x*Yxdx. 
J \ a 2 ' a 3 a 4 / J (a — x) n 

J [jjr^xp + (^=^7 + (^=^)V ** 

«/ \(« — ma) 2 (a — ma) 3 (« — ma) 4 / 
I (a -\-bx -\- cx 2 )(b -f- 2cx)dx. 

f{a + 5a + ca 2 ) n (5 + 2ca)da. 

f {b + 2cx)dx 
J (a-\-bx-\- ca 2 ) n * 



* The question whether to introduce a new symbol for a function 
whose differential is to be used must be decided by the student in each 
case. He is advised, as a rule, to first use the function, because he then 
gets a clearer view of the nature of the transformation. He can then 
replace the function by a new symbol whenever the labor of repeatedly 
writing the function will thereby be saved. 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 211 

125. Application to the Case of a Falling Body. We 

have shown (§ 33) that if, at a time t, a body is at a distance 

z from a point, the velocity of motion of the body is equal to 

dz 
the derivative — -. Now, when a body falls from a height 
at 

under the influence of a uniform force g of gravity, unmodi- 
fied by any resistance, the law in question asserts that equal 
velocities are added in equal times. That is, if z be the 
height of the body above the surface of the earth, and if we 
count the time t from the moment at which the body began 
to fall, the law asserts that 

dz . x 

the negative sign indicating that the force g acts so as to 
diminish the height z. 

By integrating this expression, we have 

z = h - §gf. (b ) 

Here the constant h represents the height z of the body at 
the moment when t = 0, or when the body began to fall. 

From the definition of h and z, it follows that h — z is the 
distance through which the body has fallen. The equation 
(b) gives 

h-z = igt\ (c) 

Hence: The distance through which the body has fallen is 
proportional to the square of the time. 

At the end of the time t the velocity of the body, meas- 
ured downwards, is, by (a), equal to gt. If at this moment 
the velocity became constant, the body would, in another 
equal interval t, move through the space gt X t = gt 9 . 
Hence, by comparing with (c) we reach by another method a 
result of §33, namely: 

In any period of time a body falls from a slate of rest 
through half the distance through which it would move in the 
same period with its acquired velocity at the end of the period. 



212 THE INTEGRAL CALCULUS. 

126. Reduction to the Logarithmic Form. Let us have 
to find 



•-/ 



771 dx 

ax -J- b' 



Since dx = —d(ax) = —d(ax + Z>), 

we may write this expression in the form 

/m d(ax -f- o) 
a ax -\-b y 

and the integral becomes 

m pdiax 4- b) m , az 4- J 

n— - -+ p-r-^ = - log , 

ajax-\~b a ° c 

c being an arbitrary constant. 

EXERCISES. 

Integrate the following expressions multiplied by dx: 
,1 b m 

I. X A . 2. — . 3. — . 

2 3 Z 

1 1 ' m 

5- kz — T- 6 - 



^ x + 1" J " 2a; - I' ' cz - 5* 

7- — + *"• 8. — - -— . 9 . 



wz ' 2a2; -{- b" %bx -f- a 8 ' 

a* + &b a + b m — n 

IO. ■ - II. — . 12. . 

—,. - p xdx p x dx 

i 3 . ita*/—? 14. yrqr^- 

Note that ».<fo == i^(a; 2 ) = |rf(l + a; 2 ). 

/jc dx , p x* dx /"log x dx 

dx 

Note that log x — = log se^ . log a:. 

a 

/»log(l + #) 7 p x dx p x dx 

18. / . , ~ dy. 19. / -- — =. 20. / -. 

J \ + y * J Va? + x* J (1 - z') f 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 213 

12*7. Trigonometric Forms, The following are examples 
of the reduction of certain trigonometric forms: 

/cos mx dx = — I cos mx d(mx) = — sin mx 4- h, 
mj v ' m 

/sin mx dx = — / sin w?£ d(vix) — li cos m#. 
Wit/ w 

/ cos (a -j- mx)dx = — / cos (a -j- mx)d(a + wa;) 

_ sin (« + w ^) , t, 
m 

//*sin a; dx pd'cosx 
tan xdx = I = — / 
t/ cos z e/ cos # 

= h — log cos a; = log c sec #, 

where ^ = log c. 

In the same way, 

/ cot xdx = log c sin a;. 

/dx /» 1 $c /^*tan z , 
- = / 7 ,- = / —r = log c tan x. 
sin z cos z J tan jc cos # </ tan x 

/dx 1 p dx 1.1 

- = - / -. — : r- = log c tan -x. 
sin x 2 J sin \x cos \x 2 

/dx p dx _ , (7t x\ 

cos x ~~ J sm (in — x) ~ ° \4 2/ 



EXERCISES. 

Integrate : 
i. (1 + cos 2/)dy. 2. (1 — e sin w)tf?«. 

3. cos 2y dy. Ans. if cos 2yd (2y) = £ sin 2y. 

4. sin 2# tfy. 5. cos ray dfy. 

6. sin y cos «/ dy. Ans. £ I sin 2yd(2y) = — J cos 2y. 

7. tan 2z cfc. 8. cot 2z *fc. 
9. 2 cos 1 a <fo. 

^4ws. / (1 + cos 2.c)rfa = a; -f- /^ + ^ sin 2x. 



214 THE INTEGRAL CALCULUS. 



IO. 


2 sin 2 x dx. 






11. 


tan 2y dy. 


12. 


cos «/ dy 
1 -f- sin y" 


^4w& J 


rrf(l + sin y) _ ,_ ,,„ , ^ _, 


1 + sin 


— ™& ^v- 1 — */• 


13- 


sin y dy 

1 -|- cos y 






14. 


sin y ft"?/ 
1 — cos y 


15- 


cos y ft*?/ 
1 — sin y 






16. 


sin %y dy 
cos ?/ 


17. 


cos 2 x — sin 

sin 2a; 


*x 


dx. 


18. 


sin 2x 

■ j r- 1 — ftZ. 

cos a; — sin x 




dx 
cos rare* 




20. 


ftz 
sin ??ia;* 


dx 


19. 


sin ??i£ cos ma; 


22. 


sin (?m# -}- a 


)dx. 


2 3- 


cos (ft — nx)dx. 


24. 


tan ?*a; ft'a;. 






25- 


tan (2a; — ft) da;. 


26. 


dx 
sin {a — x)' 




27. 


ft*a; 


„Q ^ 


cos (b — nx)' ' sin (a — nx)' 


29. 


cos «yftV 
« + sin wy" 




3°- 


sin nydy sec 2 a^a; 
« — cos ?iy ft — m tan a; 



da; da; 
128. Integration of 2 , -5 ft^d -5 5 . 

ft "J - «k" ^ — a; 

We see at once that the first differential may be reduced to 
that of an inverse tangent ; thus, 



dx 1 dx 



,-© 



Hence 



x 3 4- ft 2 ft 2 x 2 ft a; 2 

"a ~~ T" -*- a i~ -I 

ft ft 



^ 



/* *_ = L /■_*!_■ = Ltan«- «»*+*. (1) 

</ ft 2 4- a; 2 ft «/ a; 2 ft ft ' w 

•> 1 ■*■ 



r + 



We find in the same way 



,/ft 2 



— % = - tan h<-»- 4- A = ^ log c -^, (2' 
— a; 2 ft ft 2a a — x 



being an arbitrary constant factor. 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 215 
129. Integrals of the form I — — -7 — — — -^ 

fj Cl ~\~ OX ~y~ ex 

The reduction of integrals of this form depends upon the 
character of the roots of the quadratic equation 

ex* + Ix + a = 0. (1) 

I. If these roots are imaginary, the integral is the inverse 
of a trigonometric tangent. 

II. If the roots are real and unequal, the integral is the 
inverse of an hyperbolic tangent. 

III. If the roots are real and equal, that is, if the above ex- 
pression is a perfect square, the integral is an algebraic frac- 
tion. 

Dividing the denominator of the fraction by the coefficient 
of x 2 , the given integral may be written 

(a) 



-j 



bx a 

% H h - 

c c 

Writing 2p for — and q for — , the expression to be inte- 

c c 

grated may be reduced to one of the forms of § 128, thus: 

dx _ dx _ d(x-\-p) . . 



x* + 2px + q (x+py + q-p* (x + p)* + g -tf 

The three cases now depend on the sign of q — p*. 

I. If q —p* is positive, the roots of (1) are imaginary and 
the form is the first of the last article with x -f- p in the place 
of x y and q — p* in the place of a 9 . Hence we have 

dx /» d(x -f- p) 



y* ax /» 

x* + 2ps + q " J 



x* + 2px + q J (x + pY + q —p* 

= - 1 tan^ 1 ) J L±JL- + /;. (i) 

Comparing this result with (a), we see that this integral 
may be reduced to its primitive form by changing p into 



216 THE INTEGRAL CALCULUS. 

— — and q into -. Substituting and reducing, we have 
Z c c 

dx 1 /» dx 



/dx _ 1 p 

a -\- bx -\- ex* ~ c J 



. b .a 
x +c x + o 

1 J tan<-> V%C 

c 



r c 4c 2 r c 4e 2 

tan ( 1} — -j- «■ (2) 



V4ae - /> 2 r4oc - £ 2 

II. If q —p 1 is negative, that is, if 4ac — S a is negative in 
(2), the expression (2) will contain two imaginary quantities. 
But these two quantities cancel each other, so that the ex- 
pression is always real. When q — p* is negative, we write 
(b) in the form 

d{x+p) 

The integral is now in the form (2) of § 128, and we have 

/dx _ p d(x -\- p) 

x* -f 2px + q ~ ~ J p* — q — (x + pY 



Vp* — q Vp* — q 

= * - , * kg c jgEi±£±i>. (3) 

^ Vp* — £ rj? a — § f — {pc-\-p) 

Making the same substitutions in these equations that we 
made in Case I., we find 

/dx , 2 , . » 2cz 4- 5 
— — r — ; a = h tan h ( - 1} — — — { 
a + bx + ex |V _ 4 ac V^ 2 _ kac 

1 . |/£ 8 _4rt6'4-2ca; + & / , x 



VZ> 3 - 4« c W - 4ac - (2cz+ Z>) 

III. If p* — q = 0, the expression to be integrated becomes 

dx 

We have already integrated this form and found 



(*+pT 

f dx =h- 1 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 217 

EXERCISES. 

Integrate the following expressions: 

dx dx dx 

U z* -2x- 4* 2 * (x-a)(x-Ji)' 3 * a + %bx - a? 

dx dx , dx 

5- -ZTZ I?- 6 - 



^ x* + 4x + 2' D ' x(x-a)' ' (a-x)(x-b)' 

130. Inverse Sines and Cosines as Integrals. From what 
has already been shown (§ 123, (8) and (9)), it will be seen that 
we have the two following integral forms: 

— = sin (_1) x 4- h = u; (a) 

f _^_ = cos <-»z + h' = u'; (b) 

J Vl - x 2 W 

where we have added h and W as arbitrary constants of in- 
tegration. 

Comparing the first members of these equations, we see 
that each is the negative of the other. The question may 
therefore be asked why we should not write the second 
equation in the form 

u > - _ f —^— = li"- sin*"" x, (c) 

as well as in the form (b). The answer is that no error 
would arise in doing so, because the forms (b) and (c) are 
equivalent. From (b) we derive 

x = cos (u' — h') = cos {Ji f — u f )\ (d) 

and from (c), 

x = sin (h" — u'). (e) 

Now, we always have sin (a -j- 90°) = cos a. Hence (d) 
and (e) become identical by putting 

h" = V + 90°, 

which we may always do, because the value of li" is quite 
arbitrary. 



218 THE INTEGRAL CALCULUS. 

131. The preceding reasoning illustrates the fact that 
integrals expressed by circular functions may be expressed 
either in the direct or inverse form. That is, if the relation 
between the differentials of u and x is expressed in the form 

, dx 

au = — - — - — 



we may express the relation between u and x themselves 
either in the form 

u = sin (_ 1} x -f h 
or in the form x == sin (u — h). 

So, also, in the form (1) of § 128 we may express the rela- 
tion between x and u either as it is there written or in the 
reverse form, 

x = a tan a(u — h). 

dx 
132. Integration of 



We have 



Va? =F x> 



d' — 



f-^= = f-^= = sin<"« X - + A. (1) 

J Va*-x 2 J i / 1 _ x] a^ K J 

a 2 
In the same way 

f ~ dx = cos <-*>- + £ or h - sin <-»-. (2) 

We also have 

d'~ 

J Va' + z' J i/7TZ a 



= log-(* + <V + «').(8) 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 219 



d'- 

f^fi== f-=^= = cosh <-»*+ h 
J Yx' -a' J Jx] _ 1 a 



= logUx + Y^-a'). (4) 



EXERCISES. 

Integrate the differentials: 

dx dy 



3- 



Vc - x 2 
ndy 
Va 2 — ny 



mdz 

5. — = . 6. 



1/4« 2 - mV 
dx 



7< 



V±c 2 + a; 2 



<7y 

o. — • . 10. 

t^a 2 + 9*/ 2 

dy 
11. - 12. 

V* + c(z - ay 

2xdx 
13 



V4« 


- - f 

dx 




*V 


dz 


af 


V±a 


2 - m 2 z 2 




mdx 




Va 2 


dx 




Va 2 


-\- vi 2 (x 
dx 


-a)' 


V(x 


- a y- 


•4c J 


nx n 


~ 1 dx 





Va* — x* Va 2n - x 2n 

T . — cos xdx 

15. If au — —====== then sin z — a cos (?* 4- A). 

Va — sin z v 1 / 

e x dx dx 

16. - 17. 



Vl - e 2x x Vl - (log xy 

_ — sin ^efa; cos .rrf.r 

Id. -5—- — . IO. — — — — . 

fl + cos a <r -|- sin x 

(x — ^Wa; (x 4- aWfo 

20. — v ' . 21. v 

Vl - (3 - «) 4 Vl + (3 + a) 4 



220 THE INTEGRAL CALCULUS. 

dx 

133. Integration of — — — . Every differential of 

V a + bx ± ex* 

this form can be reduced to one of the three forms of the 

preceding article by a process similar to that of § 129. The 

mode of reduction will depend upon the sign of the term ex*. 

Case I. The term ex 7 is negative. Putting, as before, 

lb a 

we have 



Va + bx — ex' — Vc~ Vq + 2px — x* = V~c Vp* + q— (x-p)*. 
Then, comparing with (1) of § 132, we find 

r dx _ _i r d(x—p) 

J Va~+~bx - ex* " Ve^ Vp* -f q — (x - p)* 

1 . , n x — p 1 . 2cx — b /1X 

= —7= sm (_ 1} J = —= sm (_ 1} , (1) 

Vc Vp' + q Vc Vb* + 4ac 

In order that this expression may be real, p* -f- q or b* -\- 4=ae 
must be positive. If this quantity is negative the integral 
will be wholly imaginary, but may be reduced to an inverse 
hyperbolic sine multiplied by the imaginary unit. 

Case II. The term ex* is positive. We now have 



Va + bx -f ex* = Vc V(x + p) 2 + q — p 1 - 

dx _ 1 /» ^(a; + ?0 

Va + && + cz 2 " Vc^ V(z-\-p) 2 + q - p 2 



= —log tf(s +p + Vz> + 2ps + q) 

Vc 

i ^ 

= — log -^(2c£ + J + 2c* Va + &e + ex'). 

c ^ c * 

Because (7 is an arbitrary quantity, the quotient of C by 
2c* is equally an arbitrary quantity, and may be represented 
by the single symbol C. Thus we have 

fir* 1 _ 

- = -- log C(b+2cx+2 Vc Va+bx+cx*).(2) 



/■ 



Va + bx + ex* & 



INTEGRALS REDUCIBLE TO ELEMENTARY FORMS. 221 



EXERCISES. 

Integrate: 

dy ^ dy 



Via? + ±by - f V{$ + y) (b - y) 

ydy A dy 

4- 



VS — fy* + y* VaY - by + V 

cos Odd , cos Odd 

5. ; 6. 



Vl - sin 6 - sin 3 Vl - sin + cos 2 

sin cos Odd a sin 06/0 

o. 



V4 - cos 20 - cos 2 20 Va f - b\l - cos 0) 2 

134, Exponential Forms. Using the form (11) of § 123, 
we may reduce and integrate the simplest exponential dif- 
ferentials as follows: 

a mx dx = - / a mx d(mx) = -^ h h. (1) 

mj v m log a v 

Ca x+h dx = fa x+b d(x + b)'= j^ + A. (2) 

/I /1 „ ma? + & 
«"■« + *£& = - a mx + b d(mx-\-b) = — , M. (3) 
wy ' raloga 

— rt 

m log a 



/l /» 7i — «~ ma; 
a~ mx dx = / «- m ^(— ma;) = T . (4) 
mj x m log a v 



EXERCISES. 

Integrate: 

1. e x dx. 2. Z>*<7y. 3. a v ~ 1 dy. 

4. (a + fye x dz. 5. a y ~ c dy. 6. a~ x dx. 

7. («" + «-*)<&. 8, (a w — a"*)^. 9. (a + e*)^. 

IO< ( a k*_ fl ta)<fo ri . ___ I2 . j-p-^ 

13. ^^#. 14. (1+rfo-fe 

15. {a mx + a- mx fdx. 16. Ce^xdx. 

17. je^xdx. 18. A - °^ - J W:r. 



222 THE INTEGRAL CALCULUS. 



CHAPTER III. 

INTEGRATION BY RATIONAL TRANSFORMATIONS. 

135. We have now to consider certain forms which cannot 
be reduced so simply and directly as those treated in the last 
chapter. Before passing to general methods we shall consider 
some simple cases. 

((I _L_ x) m 

I. Integration of - ^—dx. Any form of this kind, when 

x 

m is entire, may be integrated by developing the numerator 
by the binomial theorem. We then have 

(a + x) m _a m ma m ~ x x (m\a m ~ 2 x q 

?~ ~ ^ + _ ^ r ~ + \2J~x^" + etC '' 

and each term can be integrated separately. If n < m + 2, 
and entire, one of the terms of the integral will contain log x. 

x m dx 

II. Integration of -, — " ' -. We may reduce this form to 

the preceding, by introducing a new variable, z, defined by 

the equation 

z = a -j- dx. 

This gives x = — ^ — ; dx = -j-. 

Substituting these values of x and dx in the expression to 
be integrated, it becomes 

(z — a) m dz 

which may be integrated by the method of the last article. 

xdx 

III. Integration of — -— -r — — — - 2 . We reduce the denomi- 

a ~~y~ ox jt ex 



INTEGRATION BY RATIONAL TRANSFORMATIONS. 223 

nator to the form ± ( p* - q) ± (x + p)* as in § 129. Then, 
putting, for brevity, 

z = x+p, 
which gives dx = dz, 

the integration will have to be performed on an expression of 

the form 

(z — p)dx _ zdz pdz 

¥ ± z r ~~ F±T a ~ b* ± ?' 

Each of these terms may be integrated by methods already 
given (§§126, 128). 

The process is exactly the same if we have to find 

(a -f- bx)dx 



/ 



V ± (x-pY 



EXERCISES. 



Integrate: (1 1\ 9 7 

(x — aYdx \a x ) 



X X 

x 9 dx x z dx 



" (a - xY * (1 + xY 



dx A (x -f- d)dx 

\a xj 



5 * ,n 1X3 ' 6 ' (a-x) 2 



x 3 dx 



x b dx 

7 * (a' - x 2 f 



(1 _ IV 

W x* J 

xdx zdz 



9 * a* + (b - xY IO * (a + zy + (a - zf 

(y — b)dy (z — c)dz 

I I . — ■ T 2 



(y - W + {ff + W ' a>-az + z* 
(x - a)d x (y + a )dy 

3 * x(z-b)' I4 ' a*-(y+bY 

z n dz , z b dz 



(1 + «)•' (1 - *)' 



224 THE INTEGRAL CALCULUS. 

136. Reduction of Rational Fractions in general, A ra- 
tional fraction is a fraction whose numerator and denominator 
are entire functions of the variable. The general form is 

K + VF +P*%* + . . . +p m x m N 

If the degree m of the numerator exceeds the degree n of 
the denominator, we may divide the numerator by the de- 
nominator until we have a remainder of less degree than n. 
Then, if we put Q for the entire part of the quotient, and R 
for the remainder, the fraction will be reduced to 

D ~^ + i>- 
If we have to integrate this expression, then, since Q is an 
entire function of x, the differential Qdx can be integrated 

by § 119, leaving only the proper fraction y-. Now, such a 

fraction always admits of being divided into the sum of a 
series of partial fractions with constant numerators, provided 
that we can find the roots of the equation D = 0. The theory 
of this process belongs to Algebra, but we shall show by ex- 
amples how to execute it in the three principal cases which 
may arise. 

Case I. The roots of the equation D = all real and un- 
equal. Let these roots be a, /3, y . . . 6. Then, as shown 
in Algebra, we shall have 

D = (x- a)(x - j3)(x - y) . . . (x - 6). 

We then assume 

^ = -A_ + -*- + -£- + 

D x — a^x-P^x — y^''*' 

A, B 9 C, etc., being undetermined coefficients. To deter- 
mine them we reduce the fractions in the second member to 
the common denominator D, equate the sum of the numera- 
tors of the new fractions to R, and then equate the co- 
efficients of like powers of x. 



INTEGRATION BY RATIONAL TRANSFORMATIONS. 225 



As an example, let us take the fraction 

X — X 

We readily find, by solving the equation x % — x = 0, 
a; 9 — a; = x{x — l)(a; + 1). 



Assume 

x + S A 



+^+ c 



x 3 — X X X — 1 X ~\- 1 

= (^ + # + g) jg + Cg - C) % ~ A 

X % — X 

Equating the coefficients of powers of x, we have 
A + B + = 0: 
.#- (7=1: 
^4 = -3: 
whence B = 2 and C = l. Hence 

x + S _3_ 2 1 

z 3 -z~ a; + a;-l + a; + l' 

and then, by § 120, 



«/ a 



z + 3 



*.../t + ,/ s ft I+ /* 



«/ a; " «/ a; — 1 ' •/ a;-f- 
- 3 log x + 2 log (x - 1) + log (x + 1) + log C 

gg + l)(a; - l) a 
a; 3 



log 



Integrate: 

(x — l)dx 
Im x 2 - x - 6* 
xdx 

a; a + g + 1 
5 * a; 3 +a; 2 - 6a;* 

(a; 6 + 2a ; 4 ) ^ 
7 * a; 3 + 2a: a - 8a;' 
x*dx 



EXERCISES. 



x* - (a + S)a; + a6 



aA 



2, 



1* 



(a; + a; 2 )ffo 



* (a,-l)(a;+l)(a:-2)(a;+2)- 
x*dx 



x — a 



8. 



io. 



(x* + a s )<fo 



«(»-i)(» + i)(*-a) 

dx 

x* — (a + J)iB a -f ^Z>a;* 



226 THE INTEGRAL CALCULUS. 

Case II. Some of the roots equal to each other. Let the 
factor x — a appear in D to the nth power. Then, if we 
followed the process of Case L, we should find ourselves with 
more equations than unknown quantities, because the n 
fractions 



would coalesce into one. To avoid this we write the assumed 
series of fractions in the form 

_A _B F H 

(x-a) n ~^ (x-a) n - 1 ^' ' ' ^ x-a^ x-(i~^ ' 

and then we proceed to reduce to a common denominator as 
before. The coefficients A, B, etc., are now equal in num- 
ber to the terms of the equation D = 0, so that we shall have 
exactly conditions enough to determine them. 
As an example, let it be required to integrate 

x* - 5 

— i r~r dx - 

X — X — x + 1 

We have x % - x 7 - x + 1 = (x - l) 5 (x + 1). 
We then assume 

x>-5 _A__ B C 



(x - ly (x + 1) (x-iy l x-i l x + i 

_ (B + G)x* + {A - 2C)x + A-B+C 

( x -iy(x + i) 

We find, by equating and solving, 

A = -2; 
B = +2; 
C=-l. 
Hence 

x* - 5 -2 2 1 

(x - iy(x + 1) ~ (x - i) a + x - 1 x + r 



INTEGRATION BY RATIONAL TRANSFORMATIONS. 227 
The required integral is 



dx p dx 

2 



, ■ 2 log (.t - 1) - log (x + l)+ log (7 



a; 



2 . . c(x-iy 



EXERCISES. 



Integrate: 
dx 



x(x-\-iy 

x*dx 




(x - iy(x + 2) 2# ^ (a; - df(x - b) 



(a + ic)rfic (# — a;)^ 

5 * a; 2 (a; - a) 5 " ' a; 3 (£ + af{x~^h)' 

Case III. Imaginary roots. Were the preceding methods 
applied without change to the case when the equation D = 
has imaginary roots, we should have a result in an imaginary 
form, though actually the integral is real. We therefore 
modify the process as follows: 

It is shown in Algebra that imaginary roots enter an equa- 
tion in pairs, so that if x = a -J- fii (where i = V — 1) is a 
root, then x = a — f3i will be another root. To these roots 
correspond the product 

(x — a — /3i)(x — a + /3i) = (x - a)* + f3\ 

By thus combining the imaginary factors the function D 
will be divided into factors all of which are real, but some 
of which, in the case of imaginary roots, will be of the second 
degree. 

The assumed fraction corresponding to a pair of imaginary 
roots we place in the form 

A + Bx 

(* - «)■ + P" 



228 THE INTEGRAL CALCULUS. 

and then proceed to determine A and B as before by equa- 
tions of condition. We then divide the numerator A -f- Bx 
into the two parts 

A -f- Ba and B(x — a), 

the sum of which is A -f- -##• Thus we have to integrate 

r A + Ba j P B( x - a )dx 

The first term of (a) is, by methods already developed, 
A + Ba x — a 

and the second is 

& lag ((?- ay + 0% 

We therefore have, for the complete integral, 



/ 



A -\- Bx A + Ba ,_ x — a 



tan<-»: 






EXERCISES. 



«/ 2T — 1 «/ 2 — 1 



The real factors of the denominator in Ex. 1 are (x* -j- l)(x -f- l)(x — 1). 
We resolve the given fraction in the form 

A + B x C D 

x 1 1 
and find it equal to -] — -j—. -) ^ . Then the integral is found 

to be i log (z 2 + 1) + log (z 2 - 1). 

The factors of the denominator in Ex. 2 are # — 1 and a; 2 -f- x -f- 1 = 
(* + i) 2 + £. 



'" J x 3 + 1" 4 * «/ a; 3 - 



+ l)<fo? 



+ 1* • t/ x 3 - 2x + 4* 

Note that a; -f- 2 is a factor of the denominator in (4). 



INTEGRATION BY RATIONAL TRANSFORMATIONS. 229 

137. Integration by Parts. Let u and v be any two 
functions of x. We have 

d(uv) _ dv du 
dx dx dx* 

By transposing and integrating we have 

* f u i dx = uv -f v tx dx+h > (1) 

which is a general formula of the widest application, and 
should be thoroughly memorized by the student. It shows 
us that whenever the differential function to be integrated 

can be divided into two factors, one of which (-= -dx) can be 
integrated by itself, the problem can be reduced to the inte- 
gration of some new expression Iv-^-dx). 

The formula may be written and memorized in the simpler 
form 

/ udv = uv — / vdu, (2) 

it being understood that the expressions dv and du mean dif- 
ferentials with respect to the independent variable, whatever 
that may be. 

It does not follow that the new expression will be any easier 
to integrate than the original one; and when it is not, the 
method of integrating by parts will not lead us to the integral. 
The cases in which it is applicable can only be found by trial. 

The general rule embodied in the formulae (1) and (2) is 
this : 

Express the given differential as the product of one function 
into the differential of a second function. 

Then its integral will be the product of these two functions. 
minus the integral of the second function into the differential 
of the first. 



230 TEE INTEGRAL CALCULUS. 

EXAMPLES AND EXERCISES IN INTEGRATION BY PARTS, 

i. To integrate x cos xdx. 

We have cos xdx — c^-sin x. 

Therefore in (2) we have 

u = x; v = sin x; 
and the formula becomes 

/ x cos xdx = I xd' sin x — x sin x — / sin xdx 

= x sin x -f- cos x -\- h, 

which is the required expression, as we may readily prove by 
differentiation. 

Show in the same way that — 

/ x sin xdx = — x cos x -j- sin x -\- h. 

j x sec 2 xdx = x tan x — (what ?). 

/ x sin x cos xdx — — \x cos 2x -f- i sin 2x -f- h> 

I log xdx — x log x — / :r<i • log a; = x log a; — x ~\- h. 

The process in question may be applied any number of 
times in succession. For example, 

/ x* cos xdx = / x*d"smx — x* sin x — 2 j x sin xdx. 

Then, by integrating the last term by parts, which we have 
already done, 

/ x 2 cos xdx = x 2 sin x -\- 2x cos x — 2 sin x -j- 7*. 

7. In the same way, 

/ x 3 cos xdx = / x s d'sm x = x' sin x — 3 / a- 2 sin zcdte; 

/ a; 2 sin xdx = — / a; 2 ^-cos 2 = — a; 2 cos x-\-2 / x cos a;$a;. 



INTEGRATION BY RATIONAL TRANSFORMATIONS. 231 
Then, by substitution, 

/ x 3 cos xdx = (x 3 — 6x) sin x -j- (3a; 3 — 6) cos x + h. 
8. In general, 
/ x n cos £tfo = / x n d'$m x = x n sin x — n I x n ~ l sin #t?a;; 

— / x n ~ l sin #c?z = / x tl ~ 1 d'co& x 

= x n ~ 1 cos x — (n — 1) / # n ~ 2 cos zcfo; 

— / x n ~ 2 cos zefa =— # n-2 sin x -{-{n — 2) / x n ~ 3 sin zate; 

x n ~ z sin £<fe — — a: n ~ 3 cos x -\-(n — 3) / x n ~* cos £6?£. 

etc. etc. etc. 

Then, by successive substitution, we find, for the required 
integral, 

\x n — n(n— l)x n - 2 +n(n — l)(n— 2)(n— 3)x n ~ i — . . . J sin x 
-\-\nx n ~ 1 — n(n — 1) (n — 2)x n ~ 3 -j- • • • } cos x. 

g. In the same way, show that 



sin xdx = 



■x n +n(n— l)x n ~ 2 — n(n— l){n— %)(n — 3)x n ~*+. ..} cosz 
+ \nx n ~ l - n(n - 1) (n — 2)x n ~ 3 +. . .( sin .r. 



a n log a^# = — — — - / log xd' (x n + 1 ) — — — log x 

01 -j- ±fj 71 — j— JL 

1 px n+1 _ a: M + 1 , x n + 1 
— - / ax = — — - log x — t r-^7T5« 



1 -, , _x ze~ aa; . 1 



ii. fxe- ax dx=- f-xd'(e- ax ) = -- |-- /* 

t/ J a a aj 

e~ ax dx = . 



Hence xe'^dx — 



/• 



.re 



— ax /)—ax 



232 THE INTEGRAL CALCULUS. 

12. To integrate x m e~ ax dx when m is a positive integer, we 
proceed in the same way, and repeat the process until we re- 
duce the exponent of x to unity. Thus, 

/X m B ~~ ax 771 /* 
x m e-°*dx = r- - / x m - 1 e- ax dx. 
a aj 

Treating this last integral in the same way, and repeating 
the process, the integral becomes 

1 * J -j etc. 

a a a 

e~ ax 
= — 1 (a m x m -\-ma m - 1 x m - 1 +m(m— l)a m -*x m -*-{- . . ,+m!). 

13. From the result of Ex. 5 show that 

J (log xfdx = x(F -21 + 2) +h, 

where we put, for brevity, I = log x. 

14. Show that, in general, if we put 

u n =J t (log x) n dx, 
then u n = xl n — nu n _ 1 ; 

and therefore, by successive substitution, 

u n = x(l n — nl n - 1 + n(n — l)l n ~ 2 — . . . ± til) + h. 

15. Deduce (m + 1) C Px m dx — Px m + 1 -fx m + 1 dP.' 

16. Show that if fpdx = Q, 

then / Px n dx = Qx 11 — n Qx n ~ 1 dx. 

Also, if we have 

/ Qdx : - E; I Edx = S, etc., 
then 

fPx n dx = Qx n — nRx n ~ 1 + n(n — l)Sx n ~ 2 - etc. 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 233 



CHAPTER IV. 

INTEGRATION OF IRRATIONAL ALGEBRAIC 
DIFFERENTIALS. 

138. When Fractional Powers of the Independent Vari- 
able enter into the Expression. In this case we may render 
foe expression rational by reducing the exponents to their 
least common denominator, and equating the variable to a new 
variable raised to the power represented by this denominator. 

Example. If we have to integrate 

1 + z* 
-,dx, 

then, the L. C. D. of the denominators of the exponents being 
6, we substitute for x the new variable z determined by the 
equation 

x = z% 

which gives dx = 6z*dz. 

The differential expression now reduces to 

z -f- 1 
By division this reduces to 

6 (* _* + ,- + *_,-!)& + -** 6dz 



Integrating and replacing z by its equivalent, x*, we find 






6 s 6 g 6 * 6 g 6a i 

J* = 7 X ~ 5* + 4 * + 8 * - 2* - 6:K 

+ 3 log (c 1 + 1) + 6 tan<-'> x l + 7;. 



234 THE INTEGRAL CALCULUS. 

If the fractional exponent belongs to a function of x of the 
first degree, that is, of the form ax + b, we apply the same 
method by substituting the new variable for the proper root 
of this function. 

Example. To integrate 

(a -J- bxfdx 
1 + (a + bx)' 

We put (a + bx)* = z; a -f- bx = z 2 ; 

2zdz 
ax — —7—. 
b 

The expression to be integrated now becomes 



2z 2 dz _ 2. 



dz \ 
z 2 4-l]> 



J(l + z*) b\ 
of which the integral is 

2 , x 2 

b 



(z - tan ( - J) ^ + h) = ~ | (a + &aj)*-tan<-» (« + &»)*+* 1 • 



EXERCISES. 



Integrate: 

x^dx x k dx 1 — x* 

lM T+x 2 * T+?' 3 ' 



6. 



1 + x* 




1 + a- 




(a — : 


B )» 


(2x- 


■ «)*«& 



(« — a;)*^ (« — xfdx 

I + ^T^' 5- - _ ^ _ -^ 

(x + cf ( x ~c> 1 + (2x - «)* 

1 + (z - cf, 1 - (x + fl) § 

— !— i ^fe. ii. — ' — ^ 

1 + (* - *)* 1 + (x + a)* 



1+4= 



V X -. X 5 -. 

i2. — dx, 13. -dx. 

■i I i I 

1 ^ 5 

V x X s 

(x — a)* — (x — a)* 7 

14. ^ — — -<dx, 

(x - af + (« - a) 3 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 235 

139. Cases when the Given Differential contains an 
Irrational Quantity of the Form 



V a + bx -f- ex*. 

It is a fundamental theorem of the Integral Calculus that 
if we put R = any quadratic function of x, then every ex- 
pression of the form 

F{x, VR)dx, 

{F(x, V R) being a rational function of x and \/R), admits of 
integration in terms of algebraic, logarithmic, trigonometric 
or circular functions. But if R contains terms of the third 
or any higher order in x, then the integral can, in general, be 
expressed only in terms of certain higher transcendent func- 
tions know as elliptic and Abelian functions. 

We have three cases of a quadratic function of x. 

First Case : c positive. If c is positive, we may render the 
expression rational by substituting for x the variable z, de- 
termined by the equation 



Va + ox + ex 2 = Vc(x + z) ; 
. • . a + bx + ex* = ex* + 2cxz -f- cz*. 

pyr (1 

This gives x = F -^-; (a) 

a — bz + cz' 
dx =- 2 °-(b=JcW dZ '' (i) 

Va + bx + cx> = - V£=h±^ ( C ) 

By substituting the values given by (a), (b) and (c) for the 
radical, x, and efc, the expression to be integrated will become 
rational. 

Second Case : a positive and c negative. If the term in x* 
is negative while a is positive, we put 

Va ~\-bx — ex* = Va + a»B» 

We thus derive &• == — =— ; : (a) 

z -j- c * ' 



236 THE INTEGBAL CALCULUS. 



2(Vaz 2 - Vac-bz), /3A 

dx = ~ — (7+^ dz > W 

Vaz 1 — Vac — bz 



Va + bx — ex 7 = =— ; . (c) 

z* -\- c v ' 

The substitution of these expressions will render the equa- 
tion to be integrated rational. 

Third Case : a and c both negative. If the extreme terms 
of the trinomial are both negative, we find the roots of the 
quadratic equation 

— a + bx — ex 9 = 0, 
which roots we call a and ft. We then have 

— a -\- bx — ex* = c(a — x) (x — /?), 
and we introduce the new variable z by the condition 



V— a + bx — ex' = Vc(a — x) (x — /3) = Vc(x — a)z, 
which gives x = ' ; (a) 

2+1 V ' 

substitutions which will render the equation rational. 

140. We have already integrated one expression of the 

dx 
form just considered, namely, — - without ration*- 

V a + bx + ex 1 

alization. There is yet another expression which admits of 

being integrated by a very simple transformation, namely, 

dr 



r Var 1 + br — 1 

This is the polar equation of the orbit of a planet around 
the sun. To integrate it directly, we put 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 237 

1 , dx 

x =s= — : dr == s . 

r x 

We thus reduce the expression to 

dx 



J Va 



-f- bx — x* 

Proceeding as in §133, Case I., we find the value of the 1 
integral to be 



«/ y 



dr , ~ 2x — b , ^ 2 — br 



— cos (-1) — — = cos (_1) 



Var* + Ir - 1 V4a + Z> 2 r V4a 

2 - 6r 



Thus, 6 — 7t = corf-" 



rV4o+T 2 
?r being an arbitrary constant. Hence 
2-Z>r 



r Via + b* 



COS (0 — 7r). 



Solving with respect to r, we have, for the polar equation 
of the required curve, 

2 

T ~ b + |/(4a + V) cos (0 - tt) # W 

This can be readily shown to represent an ellipse. The 
polar equation of the ellipse is, when the major axis is taken 
as the base-line and the focus as the pole, 

_ g(l - e 2 ) 2 

T ~ 1 + e cos 6 ~ 2 , 2e 

a(l — e 2 ) a(l — e 2 ) 

Comparing with (a), we have 

a(l — e 2 ) = j- = parameter of ellipse = p; 

i/(±a + b*) .... 

or e — -^ — ^ = eccentricity of ellipse. 



338 THE INTEGRAL CALCULUS. 



Irrational Binomial Forms. 

141. General Theory. An irrational binomial differen- 
tial is one in the form 

(a + bx n Yx m dx, (1) 

in which m and n are integers positive or negative,, while p is 
fractional. 

To find how and when such a form may be reduced to a 
rational one, let the fraction p, reduced to its lowest terms, be 

T 

— ; and let us put 

8 



1 
y=~(a + bx n )*. 


(2) 


This will give, when raised to the rth power 


and multi- 


plied by x m dx, 




(a -f bx n ) p x m dx = x m y r dx. 


(3) 


We readily find, from (2), 




hx n = y 3 — a; 


(«) 


sy 8 - 1 dy m 
ax ~ Imx"- 1 ' 




x m y T dx = --x m - n + 1 y r + *- l dy; 





or, substituting for x its value from (a), 

o UiS n \ m—n + \ 

x-fdx = JLpL_«)— s— y+.-yj,. (4) 

This last differential will be rational if -— is an in- 

n 

teger, which will be the case if — ^— is an integer. We shall 

call this Case I. 

To find another case when the integral may be rationalized, 
let us transform the given differential (1) by dividing the bi- 
nomial by x n and multiplying the factor outside of it by x np , 
which will leave its value unchanged. It will then be 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 239 

(b + ax - n ) p x m + n *dx, (1') 

which is another differential of the same form in which n is 
changed into —-.n and m into m + w/?. Hence, by Case I., 

this form can be made rational whenever * ~ r is an 

n 

integer; that is, when — — — f- p is such. 

it 

We have, therefore, two cases of integrability, namely: 

Case I. : when — ~3— = an integer. 

n 

Case II. : when (- p = an integer. 

Remake. It will be seen that all differentials of the form 

r 

(a + bx*) 2 x m dx must belong to one of these classes, because 

— -J — is an integer when m is odd, and — ^- (- — is such 

when m is even. In this statement we assume r to be odd, 
because if it is even the original expression is rational. 

14£. If, in Case I., the integer is -f- 1> that is, if m + 1 
= n, then the expression can be integrated immediately. 
For (4) then becomes 

the integral of which, after replacing y by its value in (2), be- 
comes 

Again, if the integer in Case II. is — 1, we have 

m + 1 + np = — n, 
or m-\- np — — n — 1. 

The expression (1) reduced to the form (1') will then be 
(b + ax~ n ) p x~ n - 1 dx = — (b -f ^-") p — ^(/> + ^-~ n ). 



240 THE INTEGRAL CALCULUS. 

which is immediately integrable, and gives by simple reduc- 
tions 

yW &*■)*— +'+■«*» = o - J^+ff^,, . (6) 

143. Forms of Reduction of Irrational Binomials. Al- 
though the integrable forms can be integrated by the substi- 
tution (2), it will, in most cases, be more convenient to ap- 
ply a system of transformations by which the integrals can be 
reduced to one of the forms just considered. The objects of 
these transformations are: 

I. To replace m by m + n or m — n; 
II. To replace p by p -j- 1 or p — 1. 

144. Firstly, to replace m by m -j- ». Let us write, for 
brevity, 

X=a + bx n , 

which will give dX = onx n ~ 1 dx, 

and the given differential will be 

X p x m dx 9 
which, again, is equal to 

zy m — n + 1 ,y m — n -\- \ 

X p dX= * , ^ d(X* +1 ). 



In bn(p-\- 1) 

Integrating by parts, we have 

/ M - = TJ^RET - R7+T3 A"'^^ (a) 

Since 

X* + 1 = JT*(a + fo n ) = aX p + £X*z n , 
the last integral in the above equation is the same as 

a fx p x m ~ n dx + bj *X p x m dx, 

of which the second integral is the same as the original one. 
Making this substitution in («), and then solving the equa- 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 241 
tion so as to obtain the value of / X p x m dx, we find 

j a x ax- &( ^ + m _ | _ 1) ^ + m+1)t / -* s Ate. ^ j 

Thus the given integral is made to depend upon another in 
which the exponent of x is changed from m to m — n. 

By reversing the equation we make the given integral 
depend on one in which the exponent is increased by n. To 
do this we change m into m + n all through the equation 
(^4), thus getting 

/ A p x m+n dx=- ■ rT x- T/ — \ — ', \ 1X / X p x m dx. 

Solving with respect to the last integral, we find 
/V„ m , X p+1 x m+1 b(np+m+n+l) /» _„ . _ /7> . 

/ X p X m dx = -7 —rr ±^-j t TV"- / X p X m + n dx. (B) 

J a(m + 1) a(m -f- 1) «/ v ' 

The repeated application of (^4) and (Z?) enables us to 
make the value of the given integral depend upon other in- 
tegrals of the same form, in which 

m is replaced by m-{-n; m-\-2n; etc.; 

or by m — n; m — 2n; etc. 

145. Next, to obtain forms in which p is increased or 
diminished by unity, we express the given differential in the 
form 



X p x m dx = X p d 



( x m + 1 \ 
\m + 1/' 



Integrating by parts and substituting for dX its value 
bnx n ~ x dx, we have 

fx p x m dx = XPxm + 1 _ JH2. fx p ->x m + n dx. (b) 

Kow, we have 

„ m + n _ ^« _ *"{* ~ <*) _ Xaf «*"» 

* "" " Z> ~ b ~b' 



242 THE INTEGRAL CALCULUS. 

and therefore, by multiplying by X p ~ 1 dx, 

fx p - 1 x m+n nx = t- fx p x m dx — j f*X p - 1 x m dx. 
Substituting this value in (£), and solving as before with 
respect to / X p x m dx, we shall find 

fx p x m dx = XPxm + 1 + an P rx*-Wdz, (C) 

in which p is diminished by unity. 

If we write p -j- 1 f or p in this equation, the last integral 
will become the given one. Doing this, and then solving 
with respect to the last integral, we find 

J an(p + l) an(p + l) J 

By the repeated application of the formula (C) or (D) we 

change 

p into p — 1, p — 2, p — 3, etc., 

or p into p + 1, /? -j- 2, p + 3, etc. 

146. To see the effect of these transformations, let us 
put, in the criteria of Oases I. and II., § 141: 

I. = i, an integer. 

TT i» + 1 ., . 

II. f- J9 = % , an integer. 

Then when we apply formula (A) or (B), since we replace 
m hy m — n or m -{- n, we have, for the new integers: 

i. «±»+i =fT i. 

II. — - — — V-p = t ' =F 1. 

It is also clear that by (6") and {D) we change II. by unity. 
Thus, every time we apply formulae {A), (B), (C) or (D) 
we change one or both of these integers by unity, so that we 
may bring them to the values unity treated in § 142. 



INTEGRATION OF IRRATIONAL DIFFERENTIALS. 243 

147. Case of Failure in this Reduction. If, in an integral 
of Case II., i f is positive, we cannot change it from zero to — 1 

771 1 1 

by the formula (A) or ((7), because, when — — \- p = 0, 

we have 

m -f- 1 + np = 0, 

and the denominators in (A) and (C) then vanish. In this 
case we have to apply the substitution of §141, without try- 
ing to reduce the integral farther. 

EXAMPLES AND EXERCISES. 

i. To integrate 

(a* ± x'fdx. 

We see that if we diminish the exponent -J by unity, we 
shall reduce the integral to a known elementary form of § 132. 
So we apply (G), putting 

m = 0; n =■ 2; p = -J; a = a 2 ; b = ± 1. 

Then ((7) becomes 

We therefore have, from § 132, 
yV + *•)*<& = | { *(«" + * ! )* + * log ^(*+(« ! + x 1 )*) | ; 

/V - «■)**! = \ | *(*' - as*)* + »' sin'-" - + A | . 
Deduce the following equations: 

3 . /(«■ + *?*&, = a + ^TT^- 1 . 

4. /V + *■)**-»& = li + - ( 4t^. 

•^ DO Ju 



244 THE INTEGRAL CALCULUS. 

7 . y ( a » _ a-*)*^ = fr( a * - x *f + fa* sin ^ -. 
Here apply formula (C); in the following (J.). 

8. Al - z a )Vefc - h - (| 2 + ~j)(l - a 9 ) 1 . 

9. To reduce and integrate (1 + x*)*x*dx. 

Here m = 3; n = 2; ^> = i; m + 1 = 4 = 2t&. We can therefore 
reduce the form to Case I. by a transformation of m into m — n, for 
which we may use either (a) or (A) of § 144. Using (a), we have 

y (1 + z 2 )* s»<fc = (1+ g 2) ^ - |y (1 + a 2 ) 1 xdx. 

The last integral can be immediately found, and gives for the required 
integral 

i(l + x>)& - &(1 + <*)*. (a) 

Using (A), we should find 

y<l + *»)* xMx = (1 + * 2 )* (£ - | g ), {&) 

a form to which (a) can be immediately reduced. 

The student will remark that the form (a) is reduced to (A) because 
in the former the exponent of X is increased by 1, which often makes 
the integration inconvenient. But when this increase of p does not in- 
terfere with the integration, we may use (a) more easily than (A). 

10. To reduce and integrate (1 + x*)*x*dx. 
Applying (A), we find 

y (1 + xrf x*dx = fl + jp* 4 - *y (1 + a»? x*dx. 

A second application repeats the form (b) above, thus giving 

11. Eeduce and integrate (1 -f x 1 )*x m dx, where m is any 
positive odd integer, and show that 



INTEGRATION OF IRRATIONAL FUNCTIONS. 245 

f(l + z')V» dx 

= f1 I t^K' 1 ( m - 1 ) x m ' S , (™-l)(™-S)x m -s \ 

K "*" ' W+2 (m + 2)m "^ (m+2)m(m-2) /' 

Remark. Where the student is writing a series of transformations he 
will find it convenient to put single symbols for the integral expressions 
which repeat themselves. Thus: 

fx h x m dx = (l); fx h x m - n <te=(2); etc. 

Thus the equations of reduction in the present example may be written 

( 1 ) = - m - T 2--^T2 X(2); 



m m 

etc. etc. 



12. Deduce the result 






246 THE INTEGRAL CALCULUS. 



CHAPTER V. 

INTEGRATION OF TRANSCENDENT FUNCTIONS. 

When the given differential contains trigonometric or other 
transcendent functions of the variable more complex than the 
simple forms treated in Chapter II. , no general method of 
reduction can be applied. Each case must therefore be 
studied for itself. 

148. To find the integrals 

/ e mx cos nxdx and / e mx sin nxdz. (1) 

Since we have 

e mx dx = -d(e mx ) = d[— \ 
m v \mj 

the integration by parts of these two expressions gives 

f^ cos nx . n 



/, e'"* cos nx , n /» m „ . , 
e mx cos nxdx = — / e vlx sm nxdx: 
m mj 

/™^ • 7 & mx sin nx n /» mr , 

e mx sm nx d x — / e « cos nxdx, 
m mj 



Solving these equations with respect to the two integrals 
which they contain, we find 



/e inx (m cos nx + n sm nx) 
e mx cos nxdx = — 5 , ' '-\ 
m -J- n 

/„„. . 7 e mx (?n sin nx — n cos nx) 
e mx gm nx fl x — v /. 
m + n 



(3) 



which are the required values. 

Eemakk. These integrals can also be obtained by substi- 
tuting for the sine and cosine their expressions in terms of 
imaginary exponentials, namely^ 



INTEGRATION OF TRANSCENDENT FUNCTIONS. 247 

2 cos nx = e nxi -f e ~ nxi } 

2 sin nx= i^" 3 * — a-" - ), 

and then integrating according to the method of § 134. The 
student should thus deduce the form (2) as an exercise. 

149. Integration of sin m x cos n xdx. 

This form is readily reducible to that of a binomial, and 
that in two ways. Since we have 

cos xdx = d'sin x, 
cos x = (1 — sin 2 #)*, 
we see that the integral may be written in the form 



/■ 



n — 1 



/' 



(1 — sin 2 x) 2 sin m xd'sm x; 
or, putting y = sin a, 

\\-tfyT L y ™ay. (3) 

By putting z = cos a; we should have, in the same way, 

/ m— 1 
(1 - z*) ~z n dz, (4) 

which is still of the same form, and is always integrable by 
the methods already developed in Chapter IV. 

If either m or n is a positive odd integer, then by develop- 
ing the binomial in (3) or (4) by the binomial theorem we 
shall reduce the expression to a series containing only posi- 
tive or negative powers of x, which is easily integrable. 

We can also, in any case, transform the integral so as to in- 
crease or diminish either of the exponents m and n by steps 
of two units at a time, as follows: 

sin m x cos n xdx — cos n ~ l x sin m xd' sin x 

__, ,sin w + 1 z 
= cos n l xd- 



m + 1 

Then, integrating by parts, we have 



248 THE IN1EQBAL CALCULUS, 



sin m x cos n xdx 
x sin 



/si, 

cos n_i 

= — — r / sm m + 2 a; cos n " 2 a^z. (5) 

m + 1 m + V w 

Because sin m + 2 2 = sin m x(l — cos 2 2;), the last term is 
equivalent to 



n 



-- / sin w x cos n ~ 2 x — - / sin m x cos n xdx. 

W m + \J 



m -\- 

The last of these factors is the original integral. Trans- 
posing the term containing it, we find 

(m + n) / sin m x cos" xdx = sin m + * x cos n _1 a; 

+ (w — 1) / sin m a; cos n ~ 8 2tffc, (6) 

in which the exponent of cos x is diminished by 2. 

We may in a similar way place the given differential in the 
form 

• «, 1 7 C0S n + 1 £ 

— sm m ~ l xd — — , 

n-\-l 

and then, proceeding as before, we shall find 

(m + n) I sin m x cos n zdfa; = — sin m _1 a; cos n + ■ a; 

+ (?ra — 1) / sin m-2 a; cos n zcfc, (7) 

thus diminishing the exponent of sin x by 2. 

By reversing these two equations we get forms in which 
the exponents are increased by 2. Writing n -\- 2 f or n in 
the first, and m -J- 2 for m in the second, we find 

(n -\-l) / sin m a; cos n a: Ja; = — sin m + x a; cos n + 1 x 

+ (ra + n + 2)/'sin m a; cos n+ 2 a;^; (8) 

(m -j- 1) / sin m a cos n a^a; = sin m + 1 x cos n + l a; 

-|- (ra + n + 2) /sin m + n x cos" z& (9) 



Mio) 



INTEGRATION OF TRANSCENDENT FUNCTIONS. 249 

150. Special cases of / sin m x cos n xdx. 

If m is zero and n is positive,, we derive, from (6), 

/„ 7 sin x cos n ~ : 2 , w — 1 /» . _ 7 
cos n zaz = h — / cos n ~ 3 am ; 
n n J 

y* „ ., 7 sin a cos n-3 £^ — 3 /* , 7 

cos n_x5 xdx = — / cos n ~ 4 xdx: 
n — % i n — 2j 

etc. etc. etc. 

The integral to be found will thus become that of cos xdx 
when n is odd, and that of dx, or x itself, when n is even. 
The given integral is then found by successive substitution. 

We find in the same way, from (7), 

/. m 7 cos£sin m-1 :£ . m — 1 /» . m «, 7 
sm m za# = — ■ / sm m ~ 2 xdx: 
m m J 

/. m „ 7 cos z sin "^z , m — 3 /» . m . , 
sm m ~ 2 za£ = - / sm m ~ 4: xdx: 

etc. etc. etc. 

From (8) and (9) we derive similar forms applicable to the 
case of negative exponents. 

EXERCISES. 

/ sin 3 x cos 8 xdx. Ans. \ cos 5 x — \ cos 3 x. 

I sin 2 x cos 3 xdx. Ans. J sin 3 x — \ sin 5 z. 



(ii) 



. 3 sin 2 x — 1 



/cos 3 zcfo 
sin 4 z * 3 sin 3 x 

/sin 3 £ tan 3 xdx. 5. / T — ^—dx. 

D J tan 3 2: 

/V v sin 3^«/. 7. fe x + a cos(x + b) dx. 

I e* v sin y cos ydy. 9. I e~ v cos 2 (1/ -f- a) dy. 

Derive the formulae of reduction 

Aan m xdx = -— Aan m + 2 iwfo: 

J m + 1 J 



250 THE INTEGRAL CALCULUS. 

and hence 

/ tan n xdx = / tan n ~ 2 xdx. 

These equations may be obtained independently by putting tan n x 
— tan n - 2 #(sec 2 x — 1); or they may be derived from (5). 

Hence derive the integrals: 

ii. / tan 3 xdx = i tan 9 x — log c sec x. (Of. § 127) 

i2./ tan 4 xdx = ^ tan 8 x — tan x -\- x -f- h. 

13. For all odd positive integral values of n, 

/» „ 7 tan n_1 £ tan n ~ 3 z , , , 

/ tan n xdx = - - — h • • ♦ ± 1°£ sec x. 

J 7i — l n — 3 ° 

14. When n is positive, integral and an even number, 

/, „ 7 tan n-1 a; tan n-3 £ , , . 

tan n xdx = - \- . . . ± tan x ± x. 
11 — I 11 — 3 

15. When the exponent is integral, odd and negative, 

/„ , cot n ~ 1 x , cot n - 3 X 
tan ~ n xdx = — - . . . ± log c sin x. 
n — 1 n — 3 

16. When the exponent is integral, even and negative, 

/„ 7 cot n_1 a; , cot n_3 £ 
tan~ n zaz = - - . . <± cot x =F %• 
n — 1 n — 3 

/* • 5 7 cos ^/ . 4 , 4 . „ , 4*2\ 

17. / sin xdx = — —— — I sin x -f- — sin 2 -j- ^ I. 

18. / sin 6 xdx 

cosxf . . , 5 . , , 5*3 . \ , 5'3x 

== ___^ ma; + _ sma;+ _ gina; j + __. 



19. / sin n # cos n xdx = — - / sin w 2£flfo 



cos 2.i; sin n_1 2z , »-l /» . n 9 ^ 7 

2 n + * ?i 2 n n J 



INTEGRATION OF TRANSCENDENT FUNCTIONS. 251 

dx 
151. To integrate „ . 3 — , — 5 = — = du. 

J in sm a; -\- n cos a; 

Dividing both terms of the fraction by cos 2 x, noticing that 
= d'tan x and writing t = tan a;, we find 



cos x 

dt 



J mH + ?i 2 x ' 



The integral is known to be (§ 128) 
— tan ( 1} - £, 

so that we have 

dx 1 . , „ m 



u 



: f , . , , 2 — = — tan<-» - tan g + A, (13) 

t/ wi sm a; + ?r cos a; mn n v ' 



m 



or tan a; = — tan mn(u — h) 



m 



152. Integration of 



a -\- b cos ?/* 

We reduce this form to the preceding one by the following 
trigonometric substitution: 

a = a (cos 2 \y -\- sin 2 \y)\ 

b cos y = £(cos 2 \y — sin 2 |?/); 

by which the expression reduces to the form 

2 r ^M , d4) 

J (a — b) sin 2 \y + (a + 6) cos 2 £y v 7 

which is that just integrated, when we put 

m = Va — b; 
n = Va-\- b. 
We therefore have 

f-rS = * = tan<-» V^4 tan I,, + h. (15) 

J a -\- b cos y \/ a * _ y* a + b "' v 



252 THE INTEGRAL CALCULUS. 

153. If, in the form of §151, m 2 and ii l have opposite 
signs, or if in § 152 we have b > a, imaginary quantities will 
enter into the integrals, although the latter are real. If, in 
the first form, the denominator is m 2 sin 2 x — n 2 cos 2 x, we 
shall have, instead of (12), the integral 

f—^—, = -L /•_* - 1 /* * ( § 136) 

e/ ra £ — w 2 /««/ mt — n ZnJ mt + n vo 7 






/ 



2m?i mt — n 
Hence, corresponding to (13), we have the result 

dx 1 . m tan x + n , , ,„ _. 

, .— a — = — s — log — ■ \- h. (16) 

vr sin a; — » cos sc 2mw ° w tan x — n v 7 



If, now, in § 152, b > a, we write (14) in the form 

_ 2 /* d^h[ 

J \b — a) sin 2 \y — {a + 6) cos 2 \if 

and instead of (15) we have the result 



/ 



du 7 , 1 , Vb—atwi^y+Vb + a ,_,„,. 



«+£cos?/ tV- ft 2 ' ^-ataniy- Vb + a 

154:, Integration of sin mx cos nxdx. 

Every form of this kind is readily integrated by substitut- 
ing for the products of sines and cosines their expressions in 
sines and cosines of the sums and differences of the angles. 
We have, by Trigonometry, 

sin mx cos nx = -A- sin (m -J- n)x -f- -j- sin (?/* — ?£):c. 

Hence 

/cos (w4- w).r cos (m — w)# , 7 
sm mx cos wzaa; = 777- — : r ^ r— + /* 
2(v/i + m) 2(m — w) 

We find in the same way 

/sin (m + w)# , sin (m — n)x , , 
cos mx cos w#a# = --rf — ■ — ~ — —, ^ — \- A 
2(m + n) 2(m — n) 

/sin (m + n)x , sin (wa — n)x , , 
sm mx sm warn = ^ — , — r - H T/ v — h #• 
2(m + n) 2(?n — n) 



INTEGRATION OF TRANSCENDENT FUNCTIONS. 253 

155. Integration by Development in Series. When the 
given derived function can be developed in a convergent 
series, we may find its integral by integrating each term of 
the series. Of course the integral will then be in the form 
of a series. The development of many known functions may 
thus be obtained. 

EXAMPLES AND EXERCISES. 

i. We may find / sin xdx as follows: We know that 



x x x' 
31 + 5 1 ~ 7! 



-■ + F T - » T + ' - 



. • . /sin xdx = h + 77 ~ r» + ;n — Q tc, 

J Z 4:1 bl 

which we recognize as the development of — cos x with an 
arbitrary constant h -\-l added to it. 

Of course we may find / cos xdx in the same way. 

dx 



2. To integrate 



l-\-x 



f l--=(l + x)-i = l-x + x*-xs+. 



x 

. r dx ?! x * . x * x * . / x 

/dx 
. = log (1 + x). Hence (a) is 

the development of log (1 -J- x), when we put li = log 1 = 0. 

The series (a) is divergent when x > 1. In this case we 
may form the development by the binomial theorem in de- 
scending powers of x, thus: 

(2 + l)- 1 ^^- 1 — x- 2 + x~ 3 — a- 4 + . . . . 

Hence we derive, when x > 1, 



254 THE INTEGRAL CALCTTLTJS. 

The arbitrary constant is zero because, when x is infinite, 
log (x -f- 1) — log x is infinitesimal. 

3. To find / — 1 = sin (_1) x in a series. 

J Vl-x* 

Hence 
p dx . . „ , 1 x 3 , 1-3 z 5 , 1-3-5 z 7 , 

The arbitrary constant is zero by the condition sin (_1) = 0. 
This series could be used for computing n by putting x = 

7t 

i, because -J = sin 30° = sin — . But its convergence would 

be much slower than that of some other series which give the 
value of n. 



dx 
rive the expansion 



4. From the equation / — = log (x -j- Vl -f- x*) de- 

«/ V14-2; 2 



1 X s . 1-3 a: 5 1-3-5 x 1 



5. By expanding a = 6Z'tan (_1) a, derive 

J. -J - 2/ 

tan<-» x = a; - %x* + K - K+ . ... 

Derive: 

P__dx_ _ 1 x 6 1-3 z 9 l-3-5 .a? 13 

,/ 4/f+z 4 2 5 2 '4'9 2-4-613 

z 3 , £ 5 £ 7 



6. 



7 . y tf --^=*+«T-3-+ s ^ l -^j l - I + 



DEFINITE INTEGRALS. 255 



CHAPTER VI. 

OF DEFINITE INTEGRALS. 

156, In the Differential Calculus the increment of a 
variable has been denned as the difference between two values 
of that variable. Let us then suppose u to represent any 
variable quantity whatever, and let us suppose u to pass 
through the series of values 







u , u x , 


u, ; 


. u„ . 


• • 


Then 


we 


shall have 
















Au 


= 


u 1 — 


^„; 








Au x 


— 


u t - 


■«*,; 








Au^ 


— 


u 3 — 


■u % ; 



Au n _ 1 = u n — w„_!. 

Taking the sum of all these equations, we have 

Au + Au x + Aii, + . . . + Au n _ 1 = u n — u ; 

That is, the difference between the two extreme values of a 
variable is equal to the sum of all the successive increments 
by which it passes from one of these values to the other. 

The same proposition may be shown graphically by sup- 
posing the variable to represent the distance from the left- 
hand end of a line to any point upon the line. The differ- 
ence between the lengths Au Q and Au 6 is evidently Au 
-f Au x + . . . + Au A . 

. | Au | Atf t | Au a I A« 3 I Au 4 1 .... 

A U U x U 2 U 3 W 4 « 6 

Since the proposition is true how small soever the incre- 
ments, it remains true when they are infinitesimal. 



256 



THE INTEGRAL CALCULUS. 




Fig. 47. 



157. Differential of an Area, Let P PP' beany curve 

whatever, and let us investigate the differential of the area 

swept over by the ordinate y 

XP. Let us suppose the 

foot of the ordinate to 

start from the position X , 

and move to the position 

X, During this motion 

XP sweeps over the area 

X P a PX, the magnitude 

of which will depend upon 

the distance OX, and will 

therefore be a function of x, which represents this distance. 

Let us put 

u = the area swept over; 

y = the ordinate XP. 

Then, if we assign to x the increment XX', the corre- 
sponding increment of the area will be XPP' X . Let us call 
y' the new ordinate X'P'. It is evident that we may always 
take the increment XX = Ax so small that the area XPP' X r 
shall be greater than yAx and less than xf Ax or vice versa. 
That is, if y f > y, as in the figure, we shall have 
yAx < Axe < if Ax. 

Now, 'when Ax approaches the limit zero, y' will approach 
y as its limit, so that the two extremes of this inequality yAx 
and y' Ax will approach equality. Hence, at the limit, 

du = ydx. (1) 

That is, the area u is such a function of x that its differen- 
tial is ydx, and its derivative with respect to x is y. 
From this it follows by integration that 



u = / ydx -j- h 



(2) 



is a general expression for the value of the area from any 
initial ordinate, as X P to the terminal ordinate XP. 



DEFINITE INTEGRALS. 



257 




Xo 



Fig. 48. 



158. The Conception of a Definite Integral. Suppose 
the area X P PX=u 
to be divided up into 
elementary areas, as in 
the figure. This area 
will then be made up of 
the sum of the areas of 
all the elementary rect- 
angles, plus that of the o^ 
triangles at the top of 
the several rectangles. That is, using the notation of § 156, 
we have 

u = y Ax + y x Ax x + y 9 Ax 9 + . . . + y n _ 1 Ax n _ 1 + T, 

T being the sum of the areas of the triangles; or, using the 
notation of sums, 

t = n — 1 

u = 2 y t Axi + T. 

» = 

Now, let each of the increments Ax t become infinitesimal. 
Then each of the small triangles which make up T will be- 
come an infinitesimal of the second order, and their sum T 
will become an infinitesimal of the first order. We may 
therefore write, for the area u, 



= OX a 

2 yAx = 



= ox 

2 ydx. 



u = lim. 

x=OX Q x=OX ~ 

That is, u is the limit of the sum of all the infinitesimal 
products ydx, as the foot of the ordinate XP moves from X 
to Xby infinitesimal steps each equal to dx. 

Such a sum of an infinite number of infinitesimal products 
is called a definite integral. 

The extreme values of the independent variable x, namely, 
OX =x and 0X= x x , are called the limits of integration. 

The infinitesimal increments ydx, whose sum makes up the 

definite integral, are called its elements. 

17 



258 THE INTEGRAL CALCULUS. 

159. Fundamental Theorem. The definite integral 
of a continuous function is equal to the difference between the 
values of the indefinite integral corresponding to the limits of 
integration. 

To show this let us write <p(x) for y, and let us put, for the 
indefinite integral, 

"cp(x)dx = F(x) + c. 



/■ 



Now, as already shown, this is a general expression for the 
area swept over by the ordinate y = <p(x), when counted from 
any arbitrary point determined by the constant c. If we 
count the area from X P , the area will be zero when x = x ; 
that is, we must have 

F(x ) + c = 0, 

which gives c — — F(x ). 

If we call x 1 the value of x at X, we shall have 

u = Area X P PX = F(x x ) + c = Fix,) - F(x ), (3) 

which was to be proved. 

We therefore have a double conception of a definite in- 
tegral, namely: 

(1) As a sum of infinitesimal products; 

(2) As the difference between two values of an indefinite 

integral; 
and it will be noticed that the identity of these two concep- 
tions rests on the theorem just enunciated. 

Notation. The definite integral is expressed in the same 
form as the indefinite integral, except that the limits of inte- 
gration are inserted after the sign / above and below the line; 
thus, 

r(p{x)dx 

means the integral of cp(x)dx taken between the limits x and 
x lt the first being the initial and the second the terminal limit. 



DEFINITE INTEGBALS. 259 

Example of the Identity of the Two Conceptions of a Defi- 
nite Integral. The double conception of a definite integral 
just reached is of fundamental importance, and may be 
further illustrated analytically. To take the simplest possible 
case, consider the definite integral 



/ 

i/x a 



adx, 



a being a constant. By definition this means the sum of all 
the products 

adx -f- adx + adx + . . . , 

as x increases from x to x x . The sum of all the dx's must 
be equal to x\ — x (§ 156). Hence 

a(dx -j- dx -j- dx -\- dx -f- . . . ) = a(x x — x ). 
But we have for the indefinite integral 



/ 



adx = ax\ 



and the definite integral is therefore, by the theorem, 

ax x — ax or a(x x — x ), 
as before. 

160. Differentiation of a Definite Integral with respect to 

its Limits. — Because the definite integral / yffo = ttmeans 

the sum of all the products ydx as x increases by infinitesimal 
increments from the lower limit x to the upper limit x x , or 

u = y dx -j- y'dx 4- y n dx + . . . -f y {n) dx, 

therefore, assigning an increment dx x to the terminal limit 
x x will add the infinitesimal increment y x dx x to u (see Fig. 48). 
That is, we shall have 

du = y x dx i9 or ~ = y x = 0^,). (4) 

In the same way, increasing the initial limit x by dx will 
take away from the sum the infinitesimal product y dx , so 



260 TEE INTEGRAL CALCULUS. 

that we shall have 

^ .= - V* = - 0fo). (5) 

The equations (4) and (5) give us the derivatives of the 
definite integral 

u = / 0(a:) . dx 

Jx 

with respect to its limits x 1 and z . 

161 • Examples and Exercises in finding Definite Inte- 
grals. 

The fundamental theorem gives the following rule for form- 
ing definite integrals: 

1. Form the indefinite integral. 

2. Substitute for the variable with respect to which we inte- 
grate, firstly, the upper limit of integration; secondly, the 
lower limit. 

3. Subtract the second result from the first. The difference 
will be the required definite integral. 

rx*dx = ^x t 3 — %x \ 
J xdx = i(b 2 — a 2 ). 3. J xdx — -J. 
/ sin xdx = — cos tt -f- cos = 2. 
I cos xdx — sin \n. 6. / azdz = \a(a % — V). 

pin -.90 

/ sin 2xdx. 8. / 

e/0 t/45° 



,-,90° 

cos 2xdx. 



„2tt p2n 

/ sin 2 xdx. 10. / cos xdx. 
Jo Jo 

Jf x sin xdx. 12. I z cos zdz. 
Jo 

I z 2 sin zdz. 14. I z 2 cos zdz. 
Jo Jo 



DEFINITE INTEGRALS. 261 

/*2jt /n't™ 

15. / z 3 cos 2zdz. 16. / z 1 sin 2z^2. 

17. / — . 18. / nz n dz, 

/ b dx p x3 dz 

tf^c? 2a J, 7=1- 

21. / cos aA. 22. / sin xdx. 

/»« + b nfl + X 

2 5' / (* ~~ <*jdx, 26. / #ay. 

t/ti — 6 t/a — a; 

•»1 — » /*« + c 

27. / (a; — l) 3 ^. 28. / (a; — a) (a: — c)d%. 

t/l + x J a - c 

.* y a y^ +y 6 «& 30. y_ t rf? = r 

IT 7T 

3 r. / a sin aa-ab. 32. / 2 cos (a + a;)6?a:. 

t/0 e/0 

cos (a: + y)dx = sin 2?/. 
y 

34. Show that / f(x)dx = — f f(x)dx. 

35. Deduce / e~ v dy = l. 

/GO } 

e~ ay dy = — . 
a 

/»° 

37. Deduce / e y ^?/ = 1. 

t/ — 00 
~GO 

38. Deduce / e~ y ' 2 ydy = £. 

39. Deduce ^ j-f^ = *• 

40. Deduce /-^ = |. 

41. Deduce / — ■ = 7t. 



262 



THE INTEGRAL CALCULUS. 



162. Failure of the Method when the Function becomes 
Infinite. It is to be noted that the equivalence of the two 
conceptions of a definite integral does not necessarily hold 
true unless the function y or <p(x) is continuous and finite 
between the limits of integration. As an example of the 
failure of this condition, 
consider the function 



y = 



(x - ay 



the curve representing 
which is shown in the 
margin. 
The indefinite integral is 



= / ydx = c — 




Fig. 4 



To find the value of this integral between two such limits 
as and k, Jc being any quantity OM less than a, we put 
x — and x = Jc, and take the difference as usual. Thus 
1 X h 

a 



u = 



(5) 



h a a(a — k)' 

JSTow, if we suppose h to approach a as its limit, so that 
a — Jc shall become infinitesimal, then the area u will increase 
without limit, as we readily see from the figure as well as by 
the formula. 

But suppose h > a; for example, Jc = 2a, Then the 
theorem would give 

,2a 



Jo ydX 



1-1- _?. 

'o a a a 

a negative finite quantity; whereas, in reality, the area is an 
infinite quantity. 

The theorem fails because, when x — a, y becomes infinite, 
so that ydx is not then necessarily an infinitesimal, as is pre- 
supposed in the demonstration. 



DEFINITE INTEGRALS. 263 

163. Change of Variable i?i Definite Integrals, When, 
in order to integrate an expression, we introduce a new vari- 
able, we must assign to the limits of integration the values of 
the new variable which correspond to the limiting values of 
the old one. Some examples will make this clear. 

Ex. 1. Let the definite integral be 

» a dx 



x 



'o a -f- x 

Proceeding in the usual way, we find the indefinite integral 

to be log (a -j- %), whence we conclude 

/~* a dx 
X ^ T ^ = log3a-log« = log3. 

But suppose that we transformed the integral by putting 

y = a -j- x; dy = dx. 

Since, at the lower limit, x = 0, we must then have y = a for 
this limit, and when, at the upper limit, x — a, we have 
y = 2a. Hence the transformed integral is 

/ a dy 
y' 

which we find to have the same value, log 2. 

Ex. 2. u = I 8 sin x(l — cos x)dx. 

We may write the indefinite integral in the form 

/ sin xdx -j- / cos #c?(cos x). 

In the first term x is still the independent variable. But, 
as the second is written, cos x is the independent variable. 
Now, for 

x = 0, cos x = 1; 

and for x = -, cos x = 0. 

Hence, writing y for cos x, the value of u is 

u = 1 2 sin xdx -j- / ydy = 1 — ^ = -J. 



264 THE INTEGRAL CALCULUS. 

Eemakk. The variable with respect to which the integra- 
tion is performed always disappears from the definite integral, 
which is a function of the limits of integration, and of any 
quantities which may enter into the differential expression. 
Hence we may change the symbol of the variable at pleasure 
without changing the integral. Thus whatever be the form 
of the function <p, or the original meaning of the symbols x 
and y, we shall always have 

P cp(x)dz = f (j>{y)dy = f <p(y + a)dy, etc. 

t/a t/a t/0 

164. Subdivision of a Definite Integral. The following 
definitions come into use here: 

1. An even function of z is a function whose value remains 
unchanged when x changes its sign. 

2. An odd function of x is one which retains the same 
absolute value with the opposite sign when x changes its sign. 

As examples: cos x is an even, sin x an odd, function. 

Any function of x % is even; the product of any even func- 
tion into x is odd. 

It is evident, from the nature and formation of a definite 
integral, that if we have a sum of such integrals, 

pb pC pd ph 

/ (}>{x)dx + / cp(x)dx + / <p(x)dx +... + / <p(x)dx, 

t/a Jb Jc t/g 

in which the upper limit of each integral is the lower limit 
of that next following, this sum is equal to 

ph 

I <p(x)dx. 

t/a 

This theorem may often be applied to simplify the expres- 
sion of the integral in cases where the values of <p(x) repeat 
themselves. 

Theorem I. If (p(x) is an even function of x, then, what- 
ever be a, 

<t>(x)dx = 2 / cp(x)dx. 



DEFINITE INTEGRALS. 265 

Because 0(— x) = <j>(x), it follows that for every negative 
value of x between — a and the element of cp(x)dx will be the 
same as for the corresponding positive value of x. Hence the 

infinitesimal sums which make up the value of / <p(x)dx will 

J — a 

be equal to those which make up / <p{x)dx. Therefore 

/a pO pa pa 

(p(x)dx= I <p(x)dx -{- / <p{x)dx = 2 j <p(x)dx. 

Theokem II. If <p{x) is an odd function of x, then, what- 
ever be a, 

+ a 

<p(x)dx = 0. 



/ 



For in this case each element 0(— x)dx will be the negative 
of the element <p(x)dx, and thus the positive and negative 
elements will cancel each other. 



i. Show that C e~ x x n dx = C flog -j dz. 

Substitute x = log -. 

z 

2. Show that whatever be the function <p, we have 

p\it p\tr 

j 0(sin z)dz = j 0(cos xdx). 
As an example of this theorem, 



M v a + b cos 71 x 7 M*a + b sm" x 

/ —i ^T-dx = / — ^ ; 

Jo a — b cos n x Jo a — b sin" x 



dx. 



The truth of this theorem may be seen by showing that to each ele- 
ment of the one integral corresponds an equal element of the other. 
Draw two quadrants; draw a sine in one and an equal cosine in the other. 
Any function <p of the sine is equal to the corresponding function of the 
cosine. We may fill one quadrant up with sines and the other with 
cosines equal to those sines, and then the two integrals will be made up 
of equal elements. 



266 TEE INTEGBAL CALCULUS. 

To express this proof analytically, we replace x by a new variable y 
= \ic — a-, which gives sin a? = cos y; dx = — dy; and then we invert 
the limits of the transformed integral, and change y into x in accordance 
with the remark of the last article. 

IT "" 

3. Show that C /(sin x)dx = 2 A /(sin x)dx. 

4. Show that / 0(sin x) cos xdx — 0. 

5. Show that if be an odd function, then 

/ 0(cos x)dx — 0. 

6. Show that the product of two like functions, odd or 
even, is an even function, and that the product of an even 
and an odd function is an odd function. 

7. Show that when is an odd function, 0(0) = 0. 

165. Definite Integrals through Integration ~by Parts. — 
In the formula for integration by parts, namely, 



/ udv = uv — / vdu, 



let us apply the rule for finding the definite integral. To ex- 
press the result, let us put 

{uv) x and (uv) Q , the values of uv for the upper and lower 
limits of integration, respectively; 

/ x i r* x \ 

udv and / vdu, the values of the two indefinite in- 
tegrals for the upper limit, x 1 ; 

J udv and jvdu, the values of the integrals for the lower 

t/X„ t/x 

limit, x . 

We then have, by the rule of § 161, 

Jf udv = I udv — / udv 
x J Jx a 



= (uv) t — I vdu — (uv) + / vdu 
= (uv) 1 — (uv) — I vdu. 

tyXn 



DEFINITE INTEGRALS. 267 

In order to assimilate the form of this expression to that of 
a definite integral, it is common to write 

{uv)\ = (uv) 1 - (uv\, 

EXAMPLES AND EXERCISES. 

i. We have found the indefinite integral 
/ log xdx = x log x — I dx. 

If we take this integral between the limits x = and x = l, 
the term x log x will vanish at both limits, so that 

(x log x) l - (x log x) = 0. 

Hence / log xdx = — / dx = — 1 + 0=— 1. 

Jo Jo 

2. To find the definite integral, 



x 



sin m xdx. 



In the equation (11), § 150, the first term of the second 
member vanishes at both the limits x = and x = n. Hence 

J/* 71 " Til — 1 p n 
' sin™ xdx = / sin m_2 :E^. 
o m Jo 

Writing m — 2 for m, and repeating the process, we have 

/* n TYl — 3 f* n 

I $m m ~ 2 xdx = - / sin™ - *£<#£; 

Jo m — 2 Jo 

s* n vti — 5 pp 

I sin m ™ i xdx = t / sin m ~ 6 xdx: 

Jo m — 4e/ 

etc. etc. 

If m is even, we shall at length reach the form 

/ dx = n — = 7T. 

Then, by successive substitution, we shall have 



268 THE INTEGRAL CALCULUS. 

f sin -» xdx = (m - 1)( CT - 3)(m - 5) 1 ^ 

Jo w(m — 2)(m — 4) ... 2 

If m is odd, fche last integral will be i sin xdx = + 2, and 
we shall have 

/ sm xdx — /v — t ^vw iT o* 

t/o m(m — 2)(m — 4) . . .3 

3. From the equation (6) of § 149 we have, by forming the 
definite integral and dividing by m -\- n, 



/ n . m „ 7 /sm m + 1 a;cos n ~ 1 ^\ 7r 
sm m x cos n #aa; = 
\ m-\-n Jo 

- I sin w x cos n ~ 2 a^fo. 

>e/0 




ft- 1 

ra + 7i, 



Since sin 7T = sin = 0, the first term of the second 
member vanishes between the limits, and we have 

y^TT ,y^ J[ /-JIT 

/ sin m x cos n xdx = ; — / sin m x cos n ~ 2 xdx. 

Jo m + nJo 

Writing n — 2, and then w — 4, etc., in place of n 9 this 
formula becomes 



/ sin m x cos n-2 xdx = — ■ -/ sin m x cos n ~ 4 xdx 

e/o m-\-n — 2 Jo 



±71- 

n — 5 



/ sin m a; cos" -4 #$e = — ; 7 / sin m a cos n_6 xdx: 

Jo m + Ti — 4 Jo 

etc. etc. 

If n is odd, the successive applications of this substitution 
will at length lead us to the form 

sin m x cos xdx = — — (sin m + 1 n — sin m + 1 0) =0; 
m + 1 v ' 

and thus, by successive substitution, we shall find all the in- 
tegrals to be zero. 



DEFINITE INTEGRALS. 269 

If n is even, we shall be led to the form 

/ sin m xdx, 

which we have just integrated. Then, by successive substi- 
tution, we find 

P sin™ 



X COS" X 



(tt--l)(rc-3) . . . 1 

(m 

dx 



-- ^_! : , / sin™ xdx 

(m + n)(m + n — 2) . . . (m + 2)J 



4. To find r-r^^ 
Jo (a* + x*) n 

We transform the differential thus: 

/dx _ 1 Mx* -f- o? — z?) dx 

_ 1 /> 6?£ 1 /» #V:e . . 

~a 2 'J (x* + a')"" 31 ~ oV p~+<r " ^ 
Integrating the last term by parts, we have 

r x'dx _\_ r 2xdx _ 1 /» ^(z 2 -t-q 2 ) 
«/ (z 2 + « 2 ) n " 2 «/ *(«" + a; 2 )" ~%J X (x* + a 2 ) 71 

1 J '*, ,_ _1 1 a^_ 

"at/ (z 2 +a 2 ) K " 1 ~ 2(» - 1)> 2 + a 2 ) n ~ 1 



1 /» dx 

^~2(n-lW (x^ + a*)"- 1 ' 

ubstituting this value of the 

A 



2(n - ly (x* + a') 
Substituting this value of the last term in (<?), we have 

dx _ 1 x 

(x* To 3 ")" " 2a> - 1) (f+Tp 1 

+ -fl-— — -)/ 1 " 
^ « 2 \ 2(» - 1V«/ ( 



2(»- l)/t/ (a 2 + « 2 )"- 1 

Passing now to the limits, we see that the first term of the 
second member vanishes both for x = and for x = oo . We 
also have 

1 _ 2n — 3 
2(w - 1) ~ 2(w - 1)' 



270 THE INTEGRAL CALCULUS. 

Hence we have the formula of reduction 

/»" dx 2n — 3 /»» dx . . 

e/o (iB' + a 1 )*""2(»-lK«/o (a f + ^)— r () 

We can thus diminish the exponent by successive steps 
until it reaches 2. The formula (b) will then give 

J/»°° dx 1 /»" Ja; 7r 

o (x* + a 2 ) 2 "" 2aV ?T~a? " 4« 3 " 

Then, by successive substitution in the form (b), we shall 
have 

r* 3 _jfa__ _ (2m — 3)(2rc — 5) . . . 1 it 
Jo (x* + a 2 )" "" (2m - 2) (2m - 4) . . . 2'2a 2n " v W 

If in (c) we suppose a = 1, and write the second member 
in reverse order, we have 



x 



dx 1-3-5 . . . (2m — 3) tt 



(1 + O w 2-4-6 . . . (27i- 2)'2* 



5- Tofind / > -^=^= = y m . 

e/o VI — a? 2 

Let us apply to the indefinite integral the formula (A), 
§ 144. We have in this case 

a — \\ b = — 1; n = 2; p = — £. 

The formula then becomes 



r x m dx x m ~ x VI — x 2 m — 1 Px m - Z dx . 

In the same way 



Px m - 2 dx _ x m ~ s Vl — x* m — 3 rx m ~ i dx 
J 4/1 _ ^ ~ m— 2 m — 2*J 



Vl — x 1, m ~ 2 m — 2«/ 4/1 _ £ 2 

Continuing the process, we shall reduce the exponent of x 
to 1 if m is odd, or to if m is even. Then we shall have 

C — = - (1 - xy or f-jjt=, = sin<"%. (J) 

Taking the several integrals between the limits and 1, we 



DEFINITE INTEGRALS. 271 

note that in (a) the first term of the second member vanishes 
at both limits, while (b) gives 

xdx p l dx 1 

7T7T. 



p x xdx _ r* x 

Jo VT^x~*~ ' J 



We thus have, by successive substitution, 

= P l x 2n+1 dx _ 2n(2n -2)(2n-4) 2 . 

^ 2n + 1 -Jo tflZTrf ~ (2»+l)(2»-l)(2»-3) . . . 3 ; 



y^=X 



1 x 2n dx _ (2n-l)(2u-S){2n-5) . . . 1 n 
\/\Z7tf ~ 2n(2n — 2){2n -4). .". . . 2'2* 



(o) 



Let us now consider the limit toward which the ratio of 
two values of y m approaches as m increases to infinity. We 
find, from (a), 

Vm _ m — l 

a ratio of which unity is the limit. 

Next we find, -by taking the quotient of the equations (c), 

it__ {2- 4- 6 . . . (2n-2)'2>i\* y 2n 

2 ~ {3-5-7 . . . (2» - l)}\2n + l)'y 2n + x ' 

Since, when n becomes infinite, the ratio y Zn : y 2n + 1 ap- 
proaches unity as its limit, we conclude that \n may be ex- 
pressed in the form of an infinite product, thus : 

n 4 4 2 6 3 8 2 10 2 



2 3'3-5'5-7'7'9'9-H 



ad infinitum. 



This is a celebrated expression for n, known as Wallis's 
formula. It cannot practically be used for computing 7t, 
owing to the great number of factors which would have to 
be included. 



272 THE INTEGRAL CALCULUS. 



CHAPTER VII. 

SUCCESSIVE INTEGRATION. 

166. Differentiation under the Sign of Integration. Let 
us have an indefinite integral of the form 

u — I <p(a, x)dx = F(a, x), (1) 

a being any quantity whatever independent of x. It is evi- 
dent that u will in general be a function of a. We have 
now to find the differential of u with respect to a. 
The differentiation of (1) gives 

du , t . 

d 2 u _ d<p(a, x) 
dadx da 

t» d *u -r.du-r.du , , . , 

Because -= — - = x> a ^— = Arr-> we have, when we consider 
dadx dx da 

du 

-j— as a function of x (cf. § 51), 

Jdu\ = dhi dx = d<f>(a, s) fc 

Then, by integrating with respect to x, 

du 
da 

in which the second member is the same as (1), except that 
(p(a, x) is replaced by its derivative with respect to a. Hence 
we have the theorem: 

The derivative of an integral with respect to any quantity 
which enters into it is expressed oy differentiating with re- 
spect to that quantity under the sign of integration. 



=f^«*> <*> 



SUCCESSIVE INTEGRATION. 273 

16T. This theorem being proved for an indefinite inte- 
gral, we have to inquire whether it can be applied to a definite 
integral. If we take the integral (1) between the limits x 
and x lf and put u and u 1 for the corresponding values of u, 
we have, for the definite integral, 

<p{a, x)dx — F(a> x t ) — F(a, x ) = u x — U B u \ 



f 

t/Xn 



Then, by differentiation, 

du* _ dF(a, x t ) dF(a, x ) 
da da da 

Comparing (1) and (2), we have 

r d0(a,x) d ^ _ dF{a, x) t 
J da da 9 



(3) 



/ 

t/x 



whence, if x x and x are not functions of a, 

*. d<f>(a, x) dx = dF(a, xj _ dF(a, aj ^ 

' Xo da da da ' ^ ' 

Hence from (3) we have the general theorem 
Da I <P{<*, %)dx = I D a (p(a, x)dx. 

l/X t/x 

That is, the symbols of differentiation and integration with 
respect to two independent quantities may be interchanged in 
a definite integral, provided that the limits of integration are 
not functions of the quantity with respect to ivhich ive differ- 
entiate. 

If the limits x 1 and x are functions of a, we have, for the 
total derivative of u * with respect to a (§ 41), 

du^ _ fdu l \ . du* dx 1 duj dx 
da \ da ) dx x da dx da' 

By § 160 we have 

£ = «**>; 



dx 
18 



dn n l 



274 THE INTEGRAL CALCULUS. 

Thus from (3) and (4) we have 

du^ />*i d(p(a, x) . x dx x , .dx /kX 

This formula is subject to the same restriction as the 
theorem for the value of a definite integral; that is, (p(a, x) 
and its derivative ivith respect to a mast be finite and con- 
tinuous for all values of ' x between the limits of integration. 

If this condition is not fulfilled, (5) may fail. 

EXERCISES. 

Differentiate: 

i. f — - — with respect to a. Ans. — ii—r — r a . 

J X -J- OL J \X -f- (X) 

2. / (x -f- ocfdx with respect to a. Ans. n I {x-\-a) n ~ l dx. 

3. J (x? -\-xyY dx with respect toy. Ans. 2 / (x 3 -\-x*y)dx. 

4. / x*dx with respect to a. Ans. of. 

t/O 

5. / x*dx with respect to a. Ans. 8a*. 

x n dx with respect to a. Ans. =a n (2a n + 1 — 1). 

And show that we have the same results in the first three 
cases whether we integrate the differential with respect to a 
or y, or differentiate the integral. 

168. The preceding method enables us to find many 
integrals, indefinite and definite, by differentiating known 
integrals with respect to constants which enter into them. 
Thus, by differentiating with respect to a the integral 



Ce ax dx — -e ax + c, 



SUCCESSIVE INTEGRATION. 275 

we find, after adding the constants of integration, 

etc. etc. 

which leads to the same results as integration by parts, and 
is shorter. 

169. The following is an instructive application of this 
and other principles. We shall hereafter show that 

f e~ x<1 dx— Vlr. 

t/ — 00 

From this it is required to find the value of C e~ a * v * dy. 

t) — oo 

If we put 

x = ay, 

dx 
whence dy = — , 

* a 9 

the corresponding indefinite integral will be 

Now, when y — ± oo , we have also x = ± oo . Hence 

e/-oo aj-, 

By differentiating with respect to a, and simple reductions, 
we find 

and from this, 

etc. etc. 






276 THE INTEGRAL CALCULUS. 

EXERCISES. 

i. By differentiating the integrals 

ycos axdx s= — sin ax, 
a 

/sin axdx == cos ax, 

twice with respect to a, prove the formulas 

2\ . . %x 



ix* cos axdx =f 3 j sin ax + -^ cos ax\ 

n 2 • 7 /2 £ 2 \ ,2a; . 

/ x sin flLTote = I — . cos ax A — -. sin ax. 

J \a a I a 

Thence show that we have 

J y* cos ydy = (y* - 2) sin y + 2y cos y; 
ftf sin ydy = (2 - */ 2 ) cos y + 2y sin y. 

2. Prove the formulae: 

CO & IS — CO tt 

zV'tfz = -; (d) / af^"£fc = (- l) n -^. 

co & t/— co a 

3. Show that the preceding formulae are true only when a 
is positive, and find the following corresponding forms when 
a has the negative sign: 

~co 1 „«> 1 

/ e~ ax dx = — ; / xe~ ax dx = — ,: 
e/o a 7 Jo a" 

/ z 2 e- aa rfe= -,; / x*e-°*dx = ~ ; etc. 
t/O « e/o a 

4. By differentiating the form of § 132, namely, 



J (a* 



dx 



sm 



(-D 



x 



(a* - xy a' 

with respect to a, show that 
dx 



/ax 
(a* - x> 



(a 2 - xy a* (a* - z 2 )** 



SUCCESSIVE INTEGBATION. 211 

170. Double Integrals. The preceding results may be 
summed up and proved thus: Let us have an integral of the 
form 

u = J <j>(x, y)dx, (1) 

and let us consider the integral 

Judy or J[_J<P( X > y)dx\dy, 

which, for brevity, is written without brackets, thus: 

<p(x, y)dxdy. 



//■ 



This expression is called a double integral. 

Theoeem. The value of an indefinite double integral re- 
mains unchanged when we change the order of the integra- 
tions, provided that we assign suitable values to the arbitrary 
constants of integration. 

Let us put 

v = J </>{?, y)dy, 

u retaining the value (1). The theorem asserts that 

/• 

= / vdx. 



fudy =j 



Call these two quantities U and V, respectively. We then 
have, by differentiation, 

dU d'U du ., x 

dV d*V dv ^ x 

dx~= V > dtdx = dy^^ X >^' 

Therefore, because of the interchangeability of differentiations, 

d — d — 

' dx ' dx 



dy dy 

Then, by integration with respect to y> 

— - — 4-c- 
dx ~ dx 



278 THE INTEGRAL CALCULUS. 

and, by integration with respect to x, 
U= V+cx + c'. 
Putting c = and c' = 0, we have U = V, as was to be 
proved. 

171. By the process of successive integration thus indi- 
cated we obtain the value of a function of two variables when 
its second derivative is given. The problem is, having an 
equation of the form 

<K*,y), (3) 



dxdy 

where (p(x, y) is supposed to be given, to find u, as a func- 
tion of x and y. This we do by integrating first with respect 
to one of the variables, say x, which will give us the value 

dn du 

of -=— , because the first member of (2) is V x -r-. Then we in- 
dy dy 

tegrate with respect to y, and thus get u. 

As an example, let us take the equation 

d 2 u . , du 2 7 

— — — = xii. or #.-— = xy dx. 
dxdy J dy * 

Integrating with respect to x, we have 

h being a quantity independent of x, which we have common- 
ly called an arbitrary constant. But, in accordance with a 
principle already laid down (§ 118), this so-called constant 
may be any quantity independent of x, and therefore any 
function we please to take of y. 

Next, integrating (3) with respect to y, and putting 



Y 



= Jhdy, 



we find u = ix*y 3 -\- Y -\- X, 

in which X is any quantity independent of y, and so may be 
an arbitrary function of x. Moreover, since h is an entirely 
arbitrary function of y } so is Y itself, 



SUCCESSIVE INTEGRATION. 279 

The student should now prove this equation by differenti- 
ating with respect to x and y in succession. 

172. Triple and Multiple Integrals. The principles just 
developed may be extended to the case of integrals involving 
three or more independent variables. The expression 

(p{x, y, z)dxdydz 



SIS' 



means the result obtained by integrating cp(x, y, z) with re- 
spect to x, then that result with respect to y, and finally that 
result with respect to z. The final result is called a triple 
integral. 

If we call F(x, y, z) the final integral to be obtained, we 

have. 

d*F(x, y, z) . x 

dxdydz =*(*>*')•> 

and the problem is to find F(x, y, z) from this equation when 
<f>{x, y, z) is given. 

Now, I say that to any integral obtained from this equation 
we may add, as arbitrary constants, three quantities: the one 
an arbitrary function of y and z\ the second an arbitrary 
function of z and x; the third an arbitrary function of x and y. 
For, let us represent any three such functions by the symbols 

{y, z], b, x], [x, y], 
and let us find the third derivative of 

F(x, y, z) + [y, z\ + [z, x] + \x, y] = u 
with respect to x, y and z. Differentiating with respect to 
x, y and z in succession, we obtain 

du _ dF(x, y,z) d\z, x] d\x, y~\ t 
dx dx dx dx ' 

d 2 u __ d?F(x, y, z) d'[x f y] m 
dxdy ~ dxdy dxdy ' 

d*u _ d 3 F(x, y, z) m 
dxdydz ~ dxdydz ' 
an equation from which the three arbitrary functions have 
entirely disappeared. 



280 THE INTEGRAL CALCULUS. 

It is to be remarked that one or both of the variables may 
disappear from any of these arbitrary functions without chang- 
ing their character. The arbitrary function of y and z, being 
any quantity whatever that does not contain x, may or may 
not contain y or z, and so with the others. 

As an example, let it be required to find 

u = / / / (x — a)(y — b)(z — c)dxdydz. 

Integrating with respect to z, and omitting the arbitrary 
function, we have 

/ / ^(x — a)(y — b)(z — cfdxdy. 

Then integrating with respect to y, 

% =/«* - «) (?/ - W '(* - <=)'; 

which gives, by integrating with respect to x, and adding the 
arbitrary functions, 

u = i(x - aY(y - l)\z - cY + [y, *] + [*, x] + \x, y\ 
The same principle may be extended to integrals with re- 
spect to any number of variables, or to multiple integrals. 
The method may also be applied to the determination of a 
function of a single variable when the derivative of the func- 
tion of any order is given. 

EXERCISES. 

i. J J^dxdy. 2. J J{x-a){y-lYdxdy. 

3. / / I xy^z^dxdydz. 4. / / j-^-dxdydz. 

5. fff( x - a Y(v - *)(* - oydxdydz. 

6. ff{x - a)*dx\ 7. fff{* + ^dz\ 

Ans. (6). ^%(x — a)* -\- Cx -\- C , G and C" being arbitrary 
constants, 



SUCCESSIVE INTEGRATION. 281 

173. Definite Double Integrals. Let U be any function 
of x and y. By integration with respect to x, supposing y 
constant, we may form a definite integral 



t/x a 



X, 



From what has been shown in § 163, Rem., U' will be a 
function of y, x and x v We may therefore form a second 
definite integral by integrating TJ'dy between two limits y 
and y x . Thus we find an expression 



f TJ'dy = I Udxdy, 

t/y t/y t/x 



which is a definite double integral. 

The limits x Q and x x of the first integration may be con- 
stants, or they may be functions of y. 

If they are constants, the two integrations will be inter- 
changeable, as shown for indefinite double integrals. 

If they are functions of y they are not interchangeable, un- 
less we make suitable changes in the limits. 

174. Definite Triple and Multiple Integrals. A definite 
integral of any order may be formed on the plan just described. 
For example, in the definite triple integral 



n> z \ /*y* s* x i 

I fifa V> %)dxdydz 

*sz Jy Q Jx 



the limits x and x x of the first integration may be functions 
of y and z; while y and y x maybe functions of z. But z and 
z 1 will be constants. 

So, in any multiple integral, the limits of the first integra- 
tion may be constants, or they may be functions of any or all 
the other variables. And each succeeding pair of limits may 
be functions of the variable which still remain, but cannot 
be functions of those with respect to which we have already 
integrated. 



282 THE INTEGRAL CALCULUS. 

EXAMPLES AND EXERCISES. 

i. Find the values of 

/ / xy^dxdy and / / xy*dxdy. 

It will be seen that in the first form the limits of x are 
constants, and in the second, functions of y. 

First integrating with respect to x, we have for the indefi- 
nite integral 

Jxtfdx = ixy, 
and for the two definite integrals 



/ xy*dx = i<fy*, 
fxtfdx = \y\ 

Uy 



>2y 

y 

Then, integrating these two functions with respect to y, 
we have 

X t C xy * dxdy = ^X a * y * dy = ia * b9 ' 

X fy Vx y" dxd y = §X y* d y = a* 5 - 

Let us now see the effect of reversing the order of the in- 
tegrations. First integrating with respect to y, we have 

/ xy*dy = %xb 3 = U. 

Then integrating with respect to x, we have 

J Udx =f a f\a/dydx = fi 2 &% 

the same result as when we integrated in the reverse order 
between the same constant limits. 



Deduce / 2 I cos (a; -J- y)dxdy = — 2. 



SUCCESSIVE INTEGRATION. 283 

3. Deduce / / cos (x — y)dxdy = -f- 4. 

4. Deduce / / (x — a)(y — b)dxdy = i« 2 Z>\ 

5. Deduce / / (x — a)(y — b)dxdy = i(2ab—a 2 )(2ab—b'), 

6. Deduce / / (x—a){y—b)dxdy = a*b — iab 3 — \a\ 

175. Product of Tivo Definite Integrals. 

Theorem. The product of the two definite integrals 

j Xdx and / Ydy is equal to the double integral 
I 1 / ' XYdxdy, provided that neither integral contains 

t/y Jx 

the variable of the other. 

For,, by hypothesis, the integral / Xdx = £7 does not con- 

i/x 

tain y. Therefore 

U j Ydy = f UYdy = / / XYdxdy, 

J tin <sy vVo vx n 

as was to be proved. 

e ~ x dx. This integral, 

QO 

which Ave have already mentioned, is a fundamental one in 
the method of least squares, and may be obtained by the ap- 
plication of the preceding theorem. Let us put 

Jc = f + ~e~ x \lx = 2 f +C °e- x \lx = 2 /* + V *\*y.(§164) 

V — 00 t/O t/0 

Then, by the theorem, 

*.= 4 f + \-*\u r + \-*dy =* r +< ° r + v<*+»*>dxa 9 . 

i'o Jo t/0 t/0 

Let us now substitute for y a new variable t, determined 
by the condition 

y = tx. 



284 THE INTEGRAL CALCULUS. 

Since, in integrating with respect to y 3 we suppose x con- 
stant, we must noAV put 

dy = xdt. 

Also, since t is infinite when y is infinite, and zero when y is 
zero, the limits of integration for t are also zero and infinity. 
Thus we have 

c/0 t/0 

Since the limits are constants, the order of integration is 
indifferent. Let us then first integrate with respect to x. 
Since 

xdx = id'x* = d' (1 + f)x\ 

the integral with respect to x is 

V*-(i + *■)&■ = 



2(1 + f)Jo 



-(l + i 2 )a; 2 -■ - • ..* « 



2(1 + n 

Then, integrating with respect to t, 

/-*°° fit 



~Vo i+.e 

Hence / e ~ x *dx — V^t, 

e/ — co 



RECTIFICATION OF CURVES. 285 



CHAPTER VIM. 
RECTIFICATION AND QUADRATURE. 

177. The Rectification of Curves. In the older geometry 
to rectify a curve meant to find a straight line equal to it in 
length. In modern geometry it means to 
find an algebraic expression for any part of 
its length. 

Let us put s for the length of the curve 
from an arbitrary fixed point C to a vari- 
able point P. If P f be another position 
of the variable point, we shall then have fig. 50. 

As = PP'. 

If PP' becomes infinitesimal, it has already been shown 
(§ 79) that we have, in rectangular co-ordinates, 




ds = Vdx< + dtf = l/l + i^Jdx = 1/1+ i^fchj, (1) 
and, in polar co-ordinates, 



ds = Vs + (%)'ae. (2) 

If both co-ordinates, x and y, are expressed in terms of a 
third variable u, we have 

* - (£*' )'+ (**• )' ! 

The length of any part of the curve is then expressed by 



286 



THE INTEGBAL CALCULUS, 



the integral of any of these expressions taken between the 
proper limits. Thus we have 



or 



•=/i'+(i)'fv 



(3) 



In order to effect the integration it is necessary that the 
second members of (3) shall be so reduced as to contain no 
other variable than that whose differencial is written; that is, 
we must have 

ds=f(x)dx; f(y)dy; f{6)dd; or f(u)du. 

Then we take for the limits of integration the values of 
x, y, 6 or u, which correspond to the ends of the curve. 

178. Rectification of the Parabola. From the equation 

of the parabola 

y 1 — 2px 

we derive ydy — pdx. 

We shall have the simplest integration by taking y as the 
independent variable. We then have 



ds = ) 1 + (~j j %; pds = \p> + f )*dy. 
The formula (C) of § 145 gives 

J J (p a + yj 



(a) 



The method of § 132 gives 
dy 



/ 



(p'+f) 



= h-logp+log((p' + y')* + y). 



RECTIFICATION OF CURVES. 287 

Thus, putting 7i f = ip (h — logp), the indefinite integral 
of (a) is 

» = *'"+ i|(/ + y 1 )* + iP % ((> + ff + y). 

The arbitrary constant h r must be so taken that s shall 
vanish at the initial point of the parabolic arc. If we take 
the vertex as this point, we must have 5 = for y = 0. Then 

It' = - \p log p. 

We therefore have, for the length of a parabolic arc from 
the vertex to the point whose ordinate is y, 

s = lp' + f) i + tplo S {P ' + f + « . (4) 

179. Rectification of the Ellipse. The formulae for rec- 
tifying the ellipse take the simplest form when we express the 
co-ordinates in terms of the eccentric angle u; then (Analyt. 
Geom. ) 

x = a cos u; y = b sin u. 

We then have 

dx = — a sin udu; dy = b cos udu. 

Then if e is the eccentricity, so that aV = a 7 — b*, 

ds = (# 2 sin 2 u + b* cos 2 u)*du = a(l — e* cos 2 ufdu\ 



s = « / (1 — e 2 cos 2 ufdu. 



This expression can be reduced to an elliptic integral: a 
kind of function which belongs to a more advanced stage of 
the calculus than that on which we are now engaged. 

It may, however, be approximately integrated by develop- 
ment in series. We have, by the binomial theorem, 

(1 — e 2 cos 2 w)* = 1 — r-6 2 cos 2 u — x- - e* cos 4 u 

' 2 2*4 

113 , , 

— e cos u — etc. 



288 THE INTEGRAL CALCULUS. 

The terms in the second member may be separately in- 
tegrated by the formulae (6), § 149, by putting m = and 
n — 2, 4, 6, etc. We thus find 

2 I cos 2 udu == sin u cos u -f- u; 

4 / cos 4 udu — sin ^(cos 3 u -f § cos «) -f fw; 
etc. etc. etc. 

Since at one end of the major axis we have u = and at 
the other end u = 7r, we find the length of one half of the 
ellipse by integrating between the limits and n. Since 
sin u vanishes at both limits, we have 



/ 



cos 5 udu = — 7l\ 

1*3 

cos 4 udu = tt-a 7 *'* 
2-4 



1-3-5 
, C ° S " = &&**' 



We thus find by substitution that the semi-circumference 
of the ellipse may be developed in powers of the eccentricity 
with the result 

U ■ 1 . 3 4 3 2 -5 6 \ 

180. The Cycloid. The co-ordinates x and y of the cy- 
cloid are expressed in terms of the angle u through which 
the generating circle has moved by the equations (§ 80) 

x = a(u — sin u); 
y = a(l — cos u). 
Hence 

ds* = dx 2 -f- dy 1 = a?{(l — cos u) 2 + sin 2 n\du* 

= 2a 2 (1 — cos u)du* = 4a 2 sin 2 \u.du?. 

By extracting the root and integrating, 

s = h — 4=a cos \u. 



RECTIFICATION OF CURVES. 289 

If we measure the arc generated from the point where it 
meets the axis of abscissas, that is, where u = 0, we must 
have s = for n = 0. This gives 

h = 4zCi 
and s = 4^(1 — cos \u) — 8a sin 2 \u. 

This gives, for the entire length of the arc generated by 
one revolution of the generating circle, 

s = 8a; 

that is, four times the diameter of the generating circle. 

181. The A rchimedean Spiral. From the polar equation 
of this spiral (§ 82) we find 

dr — add. 

Hence ds = a (I + d 2 fdd. 

Then the indefinite integral is (§ 147, Ex. 1) 

s = I j 0(1 + ?)* + % C(6 +(1 + 0'))* } . 

If we measure from the origin we must determine the value 
of C by the condition that when = 0, then s = 0. This 
gives log C = ; . * . (7=1. 

If instead of we express the length in terms of r, the 
radius vector of the terminal point of the arc, we shall have 

1 r ( 2 , i\\ , a i (^ 2 + O 1 + r 

' = a«< fl + r, > +8 1 "* 1 — s 2 — • 

182. The Logarithmic Spiral. The equation of this 
spiral (§83) gives 

— - = ale = Ir. 
do 

Hence ds = (1 + P)W0. 

To integrate this differential with respect to we should 

first substitute for r its value in terms of 0. But it will be 
19 



290 THE INTEGRAL CALCULUS. 

better to adopt the inverse course, and express dd in terms of 
dr. We thus have 

(1 + iy 

ds = dr; 

whence s = - — -r -f- s , 

s being the value of s for the pole. 

If we put y for the constant angle between the radius 
vector and the tangent, then (§§ 90-92) Z=cot y 3 and we have 

s = r sec y -f- s . 

Between any two points of the curve whose radii-vectors 
are r and r, we have 

s = (r, - r ) sec y. 

Hence the length of an arc of the logarithmic spiral is pro- 
portional to the difference between the radii-vectors of the ex- 
tremities of the arc. 

EXERCISE. 

1. Show that the differential of the arc of the lemniscate is 



Vl-2 sin 2 

(This expression can be integrated only by elliptic func- 
tions.) 

183. The Quadrature of Plane Figures. In geometrical 
construction, to square a figure means to find a square equal 
to it in area. The operation of squaring is called quadrature. 

In analysis, quadrature means the formation of an algebraic 
expression for the area of a surface. 

In order to determine an area algebraically, the equation 
of the curve which bounds it must be given. Moreover, in 
order that the area may be completely determined by the 
bounding line, the latter must be a closed curve. 

Then whatever the form of this curve, every straight line 



QUADRATURE OF PLANE FIGURES. 



291 




must intersect it an even number of times. The simplest 
case is that in which a line paral- 
lel to the axis of Y cuts the bound- 
ary in two points. Then for 
every value of x the equation of 
the curve will give two values of y 
corresponding to ordinates termi- 
nating at P and Q. Let these 
values be y and y v xo 

Then, the infinitesimal area in- fig. si. 

eluded between two ordinates infinitely near each other will 
be 

0/i —y )dx = d<r. 

The area given by integrating this expression will be 

tsx 

in which the limits of integration are the extreme values of x 
corresponding to the points X and X x , outside of which the 
ordinate ceases to cut the curve. 

The same principle may be applied by taking (x l — % )dy 
as the element of the area. We then have 

°" = f v \ x x - x My- 

If the curve is referred to polar 
co-ordinates, let 8 and T be two 
neighboring points of the curve, 
and let us put 

r=OS; 

r'=OT; 

46 = angle SOT. 

If we draw a chord from S to T, 
the area included between this chord 
and the curve will be of the third 
order (§ 78). The area of the trian 




Fig. 



;le formed by this chord 



292 



THE INTEGRAL CALCULUS. 



and the radii vectors will be \rr f sin AS. Now let Ad be- 
come infinitesimal. OS will then approach r as its limit; 
the ratio of sin Ad to Ad itself will approach unity, and the 
area of the triangle will approach that of the sector. Thus 
we shall have, for the differential of area, 

da = \r\ld. 

If the pole is within the area enclosed by the curve, the 
total area Yfill be found by integrating this expression be- 
tween the limits 0° and 360°. Thus we have, for the total 
area, 






tie. 



184. The Parabola. As the parabola is not itself a closed 
curve, it bounds no area. But we may find the area of any 
segment cut off by a double ordinate 
MX. The equation of the curve gives, 
for the two values of y, 



y x = + v*px\ y = 

Hence 



V2px. 




dcr == VSp.x^dx. 

The indefinite integral is 

o- = f V2^x~* -f- a 

For the area from the vertex to MX we put x y = OX, and 
take the integral between the limits and x x . Calling this 
area cr lf we have 

°\ = ♦ ^2px~^ = %x x y x = ix 1 x2y 1 . 

Because 2y x = MX, it follows that the areaABMX= 2x y . 
Hence: 

Theorem. The area of a parabolic segment is two thirds 
that of its circumscribed reciaiiqle. 



QUADRATURE OF PLANE FIGURES. 293 

185. The Circle and the Ellipse. Referring the circle 
of radius a to the centre as the origin, the values of y will be 

y = ± («• - x*)K 
Hence 



f(Vx - y»¥ x = 2 yV - «')*<& 



re 



= x(a 2 - x*)i + a 3 sin ( " 1) - + h. 
v ' a 

This expression, taken between appropriate limits, will 
give the area of any portion of the circle contained between 
two ordinates. 

Taking the integral between the limits — a and + a gives, 
for the area of the circle, 

a = a* sin'- 1 ) (+ 1) - a 2 sin^ (- 1) = na\ 

The Ellipse. From the equation of the ellipse referred to 
its centre and axes, namely, 

i + ^1 
a? ^ V ' 



we find y = ± - Va? 

9 a 

The entire area will be 

, 4- a 7) /, + a 



(y 1 — y )dx = 2- / (« 2 — & a )**?# = 7ra6. 

• a &J — a 

The last integration is performed exactly as in the case of 
the circle. 

186. The Hyperbola. Since the hyperbola is not a closed 
curve, it does not by itself enclose any area. But we may 
consider any area enclosed by an hyperbola and straight linos. 

Let us first consider the area A PM contained between the 
curve, the ordinate MP, and the segment AM of the major 



294 



THE INTEGRAL CALCULUS. 



axis. The equation of the hyperbola referred to its centre 
and axes gives, for the value of 
y in terms of x, 



y = — Vx* — a 2 . 

If we put a, for the value of 
the abscissa M, then, since 
OA = a, the area AMP will be 
equal to the integral 

I f X \ x * _ a *)*dx; 




Fig. 54. 



/(^-^-l^-^-Jlogg + g-iy]; 



and for the definite integral between the limits a and x, 

\hx 

2 a 
1 

2 



Area APM 



ab, Fx . (x* M 
^- ¥ log[-+(- a -lj_. 



Now, \xy is the area of the triangle OPM; we therefore 
conclude that the second term of the expression is the area 
included between. OA, OP and the hyperbolic arc A P. 

Much simpler is the area included between the curve, one 
asymptote, and two parallels 
to the other asymptote. The 
equation of the hyperbola re- 
ferred to its asymptotes as 
axes of co-ordinates (which 
axes are oblique unless the 
hyperbola is equilateral) may 
be reduced to the form 




xy 



db 



2 sin a' 



QUADRATURE OF PLANE FIGURES. 295 

a being the angle between the axes. We readily see that the 
differential of the area is ydx X sin a instead of ydx simply. 
Hence for the area we have 

ab -, ab 

ex. 



Jy sin adx = J *~dx = ~ log 



If we take the area between the limits 031 =x n and OM 



= x lf the result will be 






ab , ab , a; 



We note that this area becomes infinite when x becomes 
zero or when x x becomes infinite, showing that the entire area 
is infinite. 

187. The Lemniscaie. The equation of this curve in 
polar co-ordinates is (§ 81) 

r 2 = a' cos W. 

It will be noted that r becomes imaginary when 6 is con- 
tained between 45° and 135°, or between 225° and 315°. 
The integral expression for the area is 



fr'dd = ia 2 Aos WdO = ±a* sin 26. 



To find the area of the right-hand loop of the curve we 
must take this integral between the limits 6 = — 45° and 6 = 
-f- 45°, for which sin 2d = — 1 and + 1. Hence 

Half area = \a?\ 
Total area = a 2 . 

Hence the area of each loop of the lemniscate is half the square 
on the semi-axis. 

188. The Cycloid. By differentiating the expression for 
the abscissa of a point of the cycloid we have 

dx = a(l — cos u)du. 
Hence 



296 THE INTEGRAL CALCULUS. 

I ydx =0? / (1— cos 2iYdtt — a 2 I (f — 2 cos u -f- 1 cos 2u)du. 

The indefinite integral is 

|w — 2 sin w + i sm Sv. 

To find the whole area we take the definite integral between 
the limits and %n. Thus we find 

Area of cycloid = 37ta 2 , 

or three times the area of the generating circle. 

EXERCISES. 

i. Show that the theorem of § 184 is true only of the pa- 
rabola. 

To do this we must find what the equation of a curve must be in order 
that the theorem may be true. The theorem is 



/ 



ydx = $xy. 

Differentiating both members, we have 

ydx = \xdy + iydx ; 

, 2 <fy _ dx 
y x' 

Then, integrating both members, 

log y 2 = log ex; . * . y 2 = ex, 

c being an arbitrary constant. This is the equation of a parabola whose 
parameter is \c. 

2. Show that the equation of a curve the ratio of whose 
area to that of the circumscribed rectangle is m : n must be 
of the form 

y m = cx n ~ m . 



CUBATURE OF VOLUMES. 



297 



CHAPTER IX. 



THE CUBATURE OF VOLUMES. 

189. General Formulae for Cubature. In the ancient 
Geometry to cute a solid meant to find the edge of a cube 
whose volume should be equal to that of the solid. In Ana- 
lytic Geometry it means to find an expression for the volume 
of a solid. 

Let us have a solid the bounding surface of which is de- 
fined by an equation between rectangular co-ordinates. Let 
the solid be cut by a 
plane PL parallel to the 
plane of YZ, and let u 
be the area of the plane 
section thus formed. If 
we now cut the solid by 
a second plane, parallel 
to PL and infinitely near 
it, that portion of the 
solid contained between 
the planes will be a slice of area u and thickness dx, dx being 
the infinitesimal distance between the planes. 

If, then, we put v for the volume of that part of the solid 
contained between any two planes parallel to YZ, we have 




Fig. 56. 



and 



dv = udx, 
v = I udx, 

t/Xa 



(1) 



x and x x being the distances of the cutting planes from the 
origin 0, 



298 



THE INTEGRAL CALCULUS. 



If we take for x and x x the extreme values of x for any 
part of the solid, the above expression will give the total vol- 
ume of the solid. 

In order to integrate (1), we must express u as a function 
of x. That is, we must find a general expression in terms of 
x for the area of any section of the solid by a plane parallel 
to that of XY. This is to be done by the equation of the 
bounding surface of the solid. 

Of course we may form the infinitesimal slices by planes 
perpendicular to the axis of Y or of Z as well as of X. 

190. The Sphere. The equation of a sphere referred to 
its centre as the origin is 

z 2 + y* + z 2 = « a - 

If we cut the sphere by a plane 
PMQ parallel to the plane of 
YZ, and having the abscissa OM 
= x, the equation of the circle of 
intersection will be 

f + z * = J _ a*. 

that is, the radius MP of the circle will be Va* — x% and its 
area will be 7t{<j? — x % ). Hence the differential of the vol- 
ume of the sphere will be 

dv = 7t(a 7 — x*)dx 9 

and the indefinite integral will be 

v — 7z(a?x — ^x 3 ) -f- C. 

The extreme limits of x for the sphere are 

x = — a and x 1 = -f a. 

Taking the integral between these limits, we have 

Volume of sphere = f 7ra 3 « 




Fig. 57. 



CXTBATTTRE OF VOLUMES. 



290 



191. Volume of Pyramid. Let the pyramid be placed 
with its vertex at the ori- 
gin, and its base parallel to 
the plane of XY. Let us 
also put h = 0Z its alti- 
tude; a, the area of its base. 
Let it be cut by a plane 
EFGH parallel to its base. 

It is shown in Geometry 
that the section EFGH is 
similar to the base, and that 
the ratio of any two homologous sides, as i^Fand^i?, is the 
same as the ratio OL : OZ. Because the areas of polygons 
are proportional to the squares of their homologous sides, 

.-.Area EFGH : Area A BCD = OL' : OZ\ 
Putting Area A BCD = a, OL = z and OZ = h, 




Fig. 58. 



Area EFG II = 



az 



The volume of the pyramid is therefore 



f. 



az 2 dz 



all. 



That is, one third the altitude into the base. 
The same formulas apply to the cone. 

192. The Ellipsoid. The equation of the ellipsoid re- 
ferred to its centre and axes is 



x 1 + k 

a? "*" V 



1, 



«, b and c being the principal semi-axes. 

If we cut the ellipsoid by the plane whose equation is 
x = x', the equation of the section will be 

£ j. ?! - 1 _ <£. 

v "*" e a- • 



300 



TEE INTEGRAL CALCULUS. 



This is the equation of an ellipse whose semi-axes are 

^i^ZT^ and -Ytf^aT. 
a a 

' ., . nlcia? — x n ) 

Hence its area is — -„ -. 

a 

Then, by integration between the limits —a and -\-a, we find 
Volume of ellipsoid = ^nabc. 

From the known expression for the area of an ellipse (nab) 
it is readily found that the volume of an elliptic cylinder cir- 
cumscribing any ellipsoid is 2ft abc. Hence we conclude: 

The volume of an ellipsoid is two thirds that of any right 
elliptic cylinder circumscribed about it. 

193. Volume of any Solid of Revolution. In ©rder that 
a solid of revolution may have a well-defined volume it must 
be generated by the revo- 
lution of a curve or un- 
broken series of straight 
or curve lines terminating 
at two points, Q and R, 
of the axis of revolution. 

As an element of the volume we take two planes infinitely 
near each other and perpendicular to the axis of revolution. 
Every such plane cuts the solid in a circle. If we place the 
origin at 0, take the axis of revolution as that of X, and let 

OM = x be the abscissa of any point P of the curve, and 

MP = y its ordinate, 
then the section of the solid through M will be a circle of ra- 
dius y, whose area will therefore be ny*. 

Hence the volume contained between two planes at distance 

dx will be 

ny^dx, 

and the volume between two sections whose abscissas are x 

and x 1 will be 

V 




Fig. 59. 



/ TTlfdx. 



(1) 



CUBATURE OF VOLUMES. 



301 



If the two co-ordinates are expressed in terms of a third 
variable u by the equations 



we have 



x = <p(u), y = tp(u), 
dx = (f)'(u)du. 



Putting u and w, for the values of u corresponding to x 
and x l9 the expression (1) for the volume will become 



V=7t f Ul [zp(ti)Y<p'(u)du. 

Ju 



(2) 



M 




The equations (1) and (2) give the volume AA'B'B gem 
erated by the revolution 
of any arc AB of the 
given curve, and of the 
ordinates MA and NB 
of the extremities of the 
arc. The limits of in- 
tegration for x are OM _a.'^_ ^b 
= x and ON = x x . To fig. eo. 
find the entire volume generated we must extend these limits 
to the points (if any) at which the curve intersects the axis of 
revolution. 



194. The Paraboloid of Revolution. The equation of the 
parabola being y* = 2px, we readily 
find from (1) a result leading to the 
following theorem, which the student 
should prove for himself: 

Theorem. The volume of a para- 
boloid of revolution is one half that 
of the circumscribed cylinder. 

195. The Volume Generated by 
the Revolution of a Cycloid around 
its Base. From the equations of the cycloid in terms of 




Fig. 01. 



302 THE INTEGRAL CALCULUS. 

the angle through which the generating circle has moved, 
we find the element of the volume to be 

dV= 7ta 2 {l — cos u) 3 du, 
Hence 

V = na? I (1 — 3 cos u + 3 cos 2 u — cos 3 u)du. 

By the method of §§ 149, 150, with simple reductions, we 
find 

/ cos 3 udu = iu + i sin 2u; 

I cos 3 udu = / (1 — sin 2 u)d.$m u = sin u — J sin 3 u 

= | sin w -f- iV sin 3w. 
We thus find, for the indefinite integral, 

V = 7ta*(%u — J 4 5 ~ sin u -f- f sin 2?* — T ^ sin 3?^). 

The total volume formed by the revolution of one arc of 
the cycloid is found by taking the integral between the limits 
u = and u = 2n. The volume thus becomes 

F=5?rV, 

from which follows the theorem: 

The volume generated by the revolution of a cycloid around 
its base is five eighths that of the circumscribed cylinder, 

196. The Hyperboloid of Revolution of Two Nappes. 
This figure is formed by the revolution of an hyperbola about 
its transverse axis. The general expression for the volume is 
found to be 

h being the arbitrary constant of integration. If we consider 
that part of the infinite solid cut off by a plane perpendicular 
to the transverse axis, we must determine h by the condition 



CUBATURE OF VOLUMES. 303 

that V shall vanish when x — a, because then the plane will 
be a tangent at the vertex of the hyperboloid, and the volume 
will become zero. This condition gives 

h = da 3 - a 3 = 2a 3 . 
Thus we have 

v = £ ( * 3 - 3 ^ + 2 ^ 3) = S ( * - a) ^ x + u) - (i) 

By the same revolution whereby the hyperbola describes an 
hyperboloid of revolution the asymptotes will describe a cone. 
Let us compare the volume just found for the hyperboloid 
with that of the asymptotic cone, cut off by the same plane 
which cuts off the hyperboloid. The equation of the generat- 
ing asymptote being 

ay = bx, 
we find for the volume of the cone 

The difference between (1) and (2) will be the volume of 
the cup-shaped solid formed by cutting the hyperboloid out 
of the cone. Calling this volume V", we find 

V" = nh\x - fa). (3) 

This is equal to the volume of a circular cylinder of which 
the diameter is the conjugate axis of the hyperbola, and the 
altitude x — fa. 

This result is intimately associated with the following 
theorem, the proof of which is quite easy: 

If a plane perpendicular to the axis of revolution cut an 
hyperbola of two nappes and its asymptotic cone, the area of 
the plane contained between the circular sections is constant 
and equal to the area of the circle whose diameter is the con- 
jugate axis. 



304 



TEE INTEGRAL CALCULUS. 



'{ 


p 


r 


o 


M 


X 



Fig. 62. 



197. Ring-shaped Solids of Revolution. If any com- 
pletely bounded plane figure APQB revolve around an axis 
OX lying in its own plane, but 
wholly outside of it, it will describe 
a ring-shaped solid. 

To investigate such a solid, let the 
ordinate MP cut the figure in the 
points Q and P, and let us put 

y^MQ; y, = MP. 

The points P and Q will describe two circles which will 
contain between them the sectional area 

*&.* - y*)- 

Taking two ordinates at the infinitesimal distance dx, the 
corresponding infinitesimal element of volume will be 

dV=Tt{y:-y;)dx. (1) 

The integral 

V= 7t f X \y? - y*)dx = 7t f X \y % + y x ) (y t - yjdx 

will express the volume of that part of the solid contained be- 
tween the two planes whose respective abscissas are x and x x . 
By taking for x and x 1 the abscissas of the extreme points 
A and B, V will express the total volume of the solid. 

198. Application to the Circular Ring. Let the figure 
AB be a circle of radius c, whose centre is at the distance b 
from the axis of revolution. Let us also put 

a = the abscissa of the centre. 

We then have 



Vx = l - V c'-(x-aY ; 
y 2 = b + \fe- (x - a y ; 

y. + yt = a»; 

y. - v> = 2 iv- (x - a y ; 



CUBATUBE OF VOLUMES. 305 



V= 4:7th f\c % ~(x- ayfdx. 

t/x a 



'X 

The limits of integration for the whole volume are 
x = a — c and x x = a + c. 

If we put 

% = x — a, 

the total volume will become 



V=±7rb f + C (c 2 -zrfdz. 



By substituting the known value of the definite integral, 
we have 

V=27T 2 bc\ 

The area of the generating circle is ntf, and the circumfer- 
ence of the circle described by its centre is %7ti. The product 
of these two quantities is %n 2 ~bc 2 . Hence: 

The volume of a circular ring is equal to the product of 
the area of its cross-section into the circumference of its central 
circle. 

EXAMPLES AND EXERCISES. 

1. Compare the cycloid with the semi-ellipse having the 
same axes as the cycloid, and show the following relations be- 
tween them: 

a. The maximum radius of curvature of the ellipse (at the 
point B) is greater than that of the cycloid in the ratio 
7T 2 : 8, or 5 : 4, nearly. 

j3. The area of the semi-ellipse is greater than that of the 
cycloid in the ratio n : 3. 

y. The volume of the ellipsoid of revolution around the 
axis OX is greater than that generated by the revolution of 
the cycloid in the ratio 16 : 15. 
20 



306 



THE INTEGRAL CALCULUS. 




P\ 



Fig. 63. 



199. Quadrature of Surfaces of 
Revolution. Let us put 

As = a small arc PQ of a curve re- 
volving round an axis OX; 
y = the distance of P from the ' 
axis OX; 

y f = the distance of Q from the 
axis OX. 

Considering As as a straight line, 
the surface generated by it will be the curved surface of the 
frustum of a cone. If we put 

A a = the area of this curved surface, we have, by Geometry, 

Aa= 7t(y + y')As. 

Now let As become infinitesimal. Then y f will approach y 
as its limit, and we shall have, for the differential of the sur- 
face, 

I 



da = %7tyds — 2?ry 



1 + 



(dyX 
\dxl 



dx. 



This expression, when integrated between the limits x and 
x l9 will give the area of that portion of the surface for which 
the co-ordinates x are contained between x and x x . 

The modifications and transformations of this formula so 
as to apply it to cases when another axis than that of Y is 
the axis of revolution, or when the equation of the curve is 
not in the form y = (p(x), can be made by the student himself. 

200. Examples of Surfaces of Revolution. The process 
of applying the general formula for da to special cases is so 
like that already followed in quadrature and cubature that 
the briefest indications will suffice to guide the student. 

Surface of the Sphere. Supposing the equation of the gen- 
erating circle to be written in the form 

x~ -f if = a\ 



SURFACES OF REVOLUTION. 307 

we shall find the differential of the surface to be 
da = %nadx. 

From this we may easily prove the following : 

Theokem I. If a sphere be cut by any number of parallel 
and equidistant planes, the curved surfaces of the spherical 
zones contained between the planes will all be equal to each 
other. 

Theorem II. The total surface of a sphere is equal to the 
product of its diameter and circumference. 

Surface generated by the Revolution of a Cycloid. We shall 
find the differential of the surface to be 

da = Srta 2 sin 3 \udu. 

By a formula found in Trigonometry,, we have 

8 sin 3 v = 6 sin v — 2 sin 3v. 

Hence, putting v = \u, 

da = 4:7ta 2 (3 sin v — sin 3v)dv. 

The whole surface is obtained by integrating between the 
limits u = and u = %n\ that is, v = and v — n. We 
thus find, for the total surface, 

a = ^-7ta\ 

Hence the theorem: 

The total surface generated by the revolution of a cycloid 
about its base is four thirds the surface of the greatest in- 
scribed sphere. 

The Paraboloid of Revolution. Taking the integral be- 
tween the limits zero and x, we have for the curved surface 

V = $fi(P* + %!>*)* -p'h 
THE END, 



ID 



